Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-9-ex-9-5/

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.5

Question 1.
(x²+xy)dy = (x²+y²)dx
Solution:

Question 2.
y’ = \(\frac { x+y }{ x }\)
Solution:
We have x.\(\frac { dy }{ dx }\) = x + y
LHS as \(\frac { dy }{ dx }\) and convert RHS as a function of \(\frac { y }{ x }\) a homogeneous differential equation substitute v for \(\frac { y }{ x }\) differential y w.r.t. x put the values of \(\frac { dy }{ dx }\) and \(\frac { y }{ x }\) in (1) variable separble equation of v and x Replace the value of v by \(\frac { y }{ x }\)

which the required general solution.

Question 3.
(x – y)dy – (x + y)dx = 0
Solution:
(x – y)dy – (x + y)dx = 0
i.e., (x – y)dy = (x + y)dx

is a homogeneous differential equation

Question 4.
(x² – y²)dx + 2xy dy = 0
Solution:
(x² – y²)dx + 2xy dy = 0
2xy dy = (x² – y²)dx

is a homogeneous differential equation

which the required general solution.

Question 5.
x²\(\frac { dy }{ dx }\) = x² – 2y² + xy
Solution:
\(\frac { dy }{ dx }\) = \(\frac{x^{2}-2 y^{2}+x y}{x^{2}}\)
\(\frac { dy }{ dx }\) = \(1-2\left(\frac{y}{x}\right)^{2}+\left(\frac{y}{x}\right)\) … (1)
is a homogeneous differential equation

Question 6.
\(xdy-ydx=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } dx\)
Solution:

Question 7.
\(\left\{ xcos\left( \frac { y }{ x } \right) +ysin\left( \frac { y }{ x } \right) \right\} ydx=\left\{ ysin\left( \frac { y }{ x } \right) -xcos\left( \frac { y }{ x } \right) \right\} xdy\)
Solution:

is a function of \(\frac { y }{ x }\)
∴ The given differential equation is homogeneous.

is the general solution of the differential equation.

Question 8.
\(x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0\)
Solution:
\(x\frac { dy }{ dx } -y+xsin\left( \frac { y }{ x } \right) =0\)
i.e., x\(\frac { dy }{ dx }\) = y – xsin\(\frac { y }{ x }\)
\(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) – sin(\(\frac { y }{ x }\)) is a homogeneous differential equation.
Put \(\frac { y }{ x }\) = v ⇒ y = vx and \(\frac { dy }{ dx }\) = v + x\(\frac { dv }{ dx }\)

is the general solution.

Question 9.
\(ydx+xlog\left( \frac { y }{ x } \right) dy-2xdy=0\)
Solution:

Question 10.
\(\left( { 1+e }^{ \frac { x }{ y } } \right) dx+{ e }^{ \frac { x }{ y } }\left( 1-\frac { x }{ y } \right) dy=0\)
Solution:

is a homogeneous differential equation.
Put v = \(\frac { x }{ y }\), ∴ x = vy
Differentiating w.r.t. y we get

\(x+y e^{\frac{x}{y}}=C\), is the general solution of the differential equation.

Question 11.
(x + y) dy+(x – y)dx = 0,y = 1 when x = 1
Solution:

Question 12.
x²dy + (xy + y²)dx = 0, y = 1 when x = 1
Solution:
x²dy + (xy + y²)dx = 0
x²dy = – (xy + y²) dx

Question 13.
\(\left[x \sin ^{2}\left(\frac{y}{x}\right)-y\right] d x+x d y=0 ; y=\frac{\pi}{2}\)
Solution:

Question 14.
\(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) + cosec\(\frac { y }{ x }\) = 0; y = 0 when x = 1
Solution:
\(\frac { dy }{ dx }\) – \(\frac { y }{ x }\) + cosec\(\frac { y }{ x }\) = 0 is a homogeneous differential equation.

Question 15.
\(2xy+{ y }^{ 2 }-{ 2x }^{ 2 }\frac { dy }{ dx } =0,y=2,when\quad x=1\)
Solution:

Question 16.
A homogeneous equation of the form \(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\) can be solved by making the substitution,
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v
Solution:
\(\frac{d x}{d y}=h\left(\frac{x}{y}\right)\) is a homogeneous equation.
Hence the substitution is \(\frac { x }{ y }\) = v or x = vy

Question 17.
Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0
(b) \((x y) d x-\left(x^{3}+y^{3}\right) d y=0\)
(c) \(\left(x^{3}+2 y^{2}\right) d x+2 x y d y=0\)
(d) \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y\)
Solution:
(d) \(y^{2} d x+\left(x^{2}-x y-y^{2}\right) d y\)

Hence it is a homogeneous differential equation.

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

Class 12 Maths Chapter 7 Miscellaneous Exercise Solutions Question 1.
\(\frac{1}{x-x^{3}}\)
Solution:
\(\int \frac{1}{x-x^{3}} d x=\int \frac{1}{x\left(1-x^{2}\right)} d x=\int \frac{1}{x(1-x)(1+x)} d x\)
Let\(\frac{1}{x(1-x)(1+x)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-x}+\frac{\mathrm{C}}{1+x}\)
1 = A(1 – x)(l + x) + Bx(l + x) + Cx(l – x) Put x = 0 in (1), we get A = 1
Put x = 1 in (1), we get 1 = 2B ∴ B = \(\frac { 1 }{ 2 }\)
Put x = – 1 in (1), we get 1 = – 2C ∴ C = \(\frac { – 1 }{ 2 }\)

Miscellaneous Exercise Chapter 7 Class 12 Question 2.
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
Solution:

Miscellaneous Chapter 7 Class 12 Question 3.
\(\frac{1}{x \sqrt{a x-x^{2}}}\)
Solution:

Ncert Solutions For Class 12 Maths Chapter 7 Miscellaneous Exercise Question 4.
\(\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}\)
Solution:

Miscellaneous Ch 7 Class 12 Question 5.
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\)
Solution:

Miscellaneous Exercise Class 12 Chapter 7 Question 6.
\(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\) = \(\frac{A}{x+1}+\frac{B x+C}{x^{2}+9}\)
⇒ 5x = A(x² + 9) + (Bx + C)(x + 1) … (1)
Put x = – 1 in (1), we get
– 5 = – 10
∴ A = \(\frac { -5 }{ 10 }\) = \(\frac { – 1 }{ 2 }\)
Equating the coefficients of x² and constant
term, we get A + B = 0, 9 A + C = 0

Ch 7 Miscellaneous Class 12 Question 7.
\(\frac{\sin x}{\sin (x-a)}\)
Solution:

= sin a ∫cot dt + cos a ∫dt
= sin a log |sin t| + cos a (t) + C₁
= sin a log |sin(x – a)| + cos a[x – 1] + C₁
= sin a.logsin (x – a) + x cosa + C, where C = C₁ – a cos a

Ncert Solutions Class 12 Maths Chapter 7 Miscellaneous Exercise Question 8.
\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
Solution:

Miscellaneous Exercise On Chapter 7 Class 12 Question 9.
\(\frac{\cos x}{\sqrt{4-\sin ^{2} x}}\)
Solution:

Integration Miscellaneous Class 12 Question 10.
\(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)
Solution:
Let I = \(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)dx
sin⁸x – cos⁸x (sin⁴x – cos⁴x)(sin⁴x + cos⁴x)
= (sin²x – cos²x)(sin²x + cos²x)(sin⁴x + cos⁴x)
= (sn²x – cos²x(1)[(sin²x + cos²x) – 2sin²x cos²x]
= (sin²x – cos²x)( 1 – 2sin²x cos²x)

Question 11.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution:

Question 12.
\(\frac{x^{3}}{\sqrt{1-x^{8}}}\)
Solution:

Question 13.
\(\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}\)
Solution:
Let I = \(\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}\) dx
Put t = e^x
\(\frac { dt }{ dx }\) = e^x
dt = e^x dx
= \(\int \frac{d t}{(1+t)(2+t)}\)
Let \(\frac{1}{(1+t)(2+t)}=\frac{\mathrm{A}}{(1+t)}+\frac{\mathrm{B}}{(2+t)}\)
∴ 1 = A(2 + t) + B(1 + t) … (1)
Put t = – 2 in (1), we get A = 1
Put t = – 2 in (1), we get 1 = – B ∴B = – 1
\(\frac{1}{(1+t)(2+t)}=\frac{1}{(1+t)}-\frac{1}{(2+t)}\)
I = \(\int \frac{1}{1+t} d t-\int \frac{1}{2+t} d t\)
= \(\log |1+t|-\log |2+t|+C\)
= \(\log \left|\frac{1+t}{2+t}\right|+\mathrm{C}=\log \left(\frac{1+e^{x}}{2+e^{x}}\right)+\mathrm{C}\)

Question 14.
\(\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}\)
Solution:

Question 15.
\(\cos ^{3} x e^{\log \sin x}\)
Solution:

Question 16.
\(e^{3 \log x}\left(x^{4}+1\right)^{-1}\)
Solution:

Question 17.
\(f^{\prime}(a x+b)[f(a x+b)]^{n}\)
Solution:

Question 18.
\(\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}\)
Solution:

Question 19.
\(\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]\)
Solution:

Question 20.
\(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\)
Solution:

Question 21.
\(\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}\)
Solution:

Question 22.
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\)
Solution:
Let \(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{\mathrm{A}}{(x+1)}+\frac{\mathrm{B}}{(x+1)^{2}}+\frac{\mathrm{C}}{(x+2)}\)
⇒ x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)² … (1)
Put x = – 1 in (1), we get B = 1
Put x = – 2 in (1), we get C = 3
Equating the coefficients of x2, we get
A + C = 1 ∴ A = – 2

Question 23.
\(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:

Question 24.
\(\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}\)
Solution:

Question 25.
\(\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\)
Solution:

Question 26.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x\)
Solution:

Question 27.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\)
Solution:
Let I = \(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\)
Dividing the Nr. and Dr. by cos²x, we get

where t² = y
⇒ 1 = A(1 + 4y) + B(1 + y) … (1)
Put y = – 1 in (1), we get 1 = – 3A ∴ A = \(\frac { – 1 }{ 3 }\)
Equating the coefficients of y, we get
4A + B = 0
∴ B = \(\frac { 4 }{ 3 }\)

Question 28.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Solution:

Question 29.
\(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}\)
Solution:

Question 30.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\)
Solution:

Question 31.
\(\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x\)
Solution:

Question 32.
\(\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x\)
Solution:

Question 33.
\(\left.\int_{1}^{4}|| x-1|+| x-2|+| x-3 \mid\right] d x\)
Solution:

Question 34.
\(\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}\)
Solution:
Let \(\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\)
1 = Ax(x + 1) + B(x + 1) + C(x²) … (1)
Put x = 0 in (1), we get B = 1
Put x = – 1 in (1), we get C = 1
Equating the coefficients of x², we get
A + C = 0 ∴ A = – 1

Question 35.
\(\int_{0}^{1} x e^{x} d x=1\)
Solution:

Question 36.
\(\int_{-1}^{1} x^{17} \cos ^{4} x d x=0\)
Solution:
Let f(x) = x¹⁷cos⁴x
f(- x) = (- x)¹⁷cos< sup>4(-x)
∴ f(x) is an odd function.
∴ \(\int_{-1}^{1} x^{17} \cos ^{4} x \cdot d x=0\)
\(\int_{-a}^{a} f(x) d x=0, \text { if } f(x) \text { is odd }\)

Question 37.
\(\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}\)
Solution:

Question 38.
\(\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2\)
Solution:

Question 39.
\(\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1\)
Solution
∫sin^-1 x dx = x sin^-1 x + \(\sqrt{1-x^{2}}\)
(Refer in-tegrals of inverse trigonometric functions)
∴ \(\int_{0}^{1} \sin ^{-1} x d x=\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]_{0}^{1}\)
= \(\left(\frac{\pi}{2}+0\right)-(0+1)=\frac{\pi}{2}-1\)

Question 40.
Evaluate \(\int_{0}^{1} e^{2-3 x} d x\) as a limit of a sum.
Solution:
Let I = \(\int_{0}^{1} e^{2-3 x} d x\)
Here f(x) = e^2-3x, a = 0, b = 1
nh = b – a = 1 – 0 = 1
f(0 + h) = f(h) = e^2-3h
f(0 + 2h) = f(2h) = e^2-6h
……………………………….
f(0 + (n – 1)h = f((n – 1)h) = e^2-3(n-1)h

Question 41.
\(\int \frac{d x}{e^{x}+e^{-x}}\) is equal to
a. tan^-1(e^x) + C
b. tan^-1(e^-x) + C
c. log(e^x – e^-x) + C
d. log(e^x + x^-x) + C
Solution:

Question 42.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x\) is equal to
a. \(\frac{-1}{\sin x+\cos x}+\mathrm{C}\)
b. log|sin x + cos x| + C
c. log|sin x – cos x| + C
d. \(\frac{1}{(\sin x+\cos x)^{2}}\)
Solution:

Question 43.
If \(f(a+b-x)=f(x), \text { then } \int_{a}^{b} x f(x) d x\) is equal to
a. \(\frac{a+b}{2} \int_{a}^{b} f(b-x) d x\)
b. \(\frac{a+b}{2} \int_{a}^{b} f(b+x) d x\)
c. \(\frac{b-a}{2} \int_{a}^{b} f(x) d x\)
d. \(\frac{a+b}{2} \int_{a}^{b} f(x) d x\)
Solution:

Question 44.
The value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x\) is
a. 1
b. 0
c. – 1
d. \(\frac{\pi}{4}\)
Solution:

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-8-ex-8-2/

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2

Ex 8.2 Class 12 NCERT Solutions Question 1.
Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y.
Solution:

4x² + 4y² = 9
∴ x² + y² = \(\frac { 9 }{ 4 }\) … (1) is a circle with
centre (0,0) and radius = \(\frac { 3 }{ 2 }\)
x² = 4y … (2)
Substituting (2) in (1), we get 4y + y²
⇒ 4y² + 16y – 9 = 0
⇒ 4y² – 2y + 18y – 9 = 0
⇒ 2y(2y – 1) + 9(2y – 1) = 0
⇒ (2y – 1)(2y + 9) = 0
⇒ y = \(\frac { 1 }{ 2 }\), y = \(\frac { – 9 }{ 2 }\)
∴ y = \(\frac { 1 }{ 2 }\)
Substitute y = \(\frac { 1 }{ 2 }\) in (2) we get x = ± \(\sqrt{2}\)
∴ The point of intersection of the curves are (- \(\sqrt{2}\), \(\frac { 1 }{ 2 }\)) and (\(\sqrt{2}\), \(\frac { 1 }{ 2 }\))
Since the shaded region is symmetric with respect to y-axis,
Required Area = 2[Area under the circle – Area under the parabola] lying in the first quadrant between the lines x = 0 and x = \(\sqrt{2}\)

Exercise 8.2 Class 12 NCERT Solutions Question 2.
Find the area bounded by curves (x – 1)² + y² = 1 and x² + y² = 1.
Solution:

x² + y² = 1 is a circle with centre at origin and intersecting positive x-axis at (1, 0).
(x – 1)² + y = 1 is a circle with centre at the origin and passing through the origin.
Solving x² + y² = 1 and (x – 1)² + y² = 1,
we get x² – (x – 1)² = 0
(x – x + 1)(x + x – 1) = 0
2x – 1 = 0
∴ x = \(\frac { 1 }{ 2 }\)
When x = \(\frac { 1 }{ 2 }\), we get (\(\frac { 1 }{ 2 }\))² + y² = 1
y² = \(\frac { 3 }{ 4 }\)
∴ y = \(\frac{\pm \sqrt{3}}{2}\)
Hence the points of intersection of the two circles are (\(\frac { 1 }{ 2 }\), \(\frac{\pm \sqrt{3}}{2}\)) and (\(\frac { 1 }{ 2 }\), \(\frac{\pm \sqrt{- 3}}{2}\)).
The area bounded by the two curves is sym-metric w.r.t. x-axis.
∴ Required area = 2 x shaded area … (1)
Area of shaded region = Area under arc OA + Area under arc AC

8.2 Class 12 NCERT Solutions Question 3.
Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 3.
Solution:
Equation of the parabola is y = x² + 2 or x² = (y – 2)
Its vertex is (0,2) axis is y-axis.
Boundary lines are y = x, x = 0, x = 3.
Graphs of the curve and lines have been shown in the figure.
Area of the region PQRO = Area of the region OAQR – Area of region OAP

Ex 8.2 Class 12 Maths Ncert Solutions Question 4.
Using integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).
Solution:
Let the vertices of the triangle are A(- 1, 0), B(1, 3) and C(3, 2).

Equation of AB is y – 0 = \(\frac { 3 – 0 }{ 1+1 }\)(x+1)
i.e, y = \(\frac { 3 }{ 2 }\)(x + 1)
Equation of BC is y – 3 = \(\frac { 2 – 3 }{ 3 – 1 }\)(x – 1)
i.e, y – 3 = \(\frac { – 1 }{ 2 }\)(x – 1) ∴ y = \(\frac { 7 – x }{ 2 }\)
Equation of AC is y – 0 = \(\frac { 2 – 0 }{ 3 + 1 }\) (x + 1)
i.e., y = \(\frac { 1 }{ 2 }\)(x + 1)
Area of ∆ ABC = Area of ∆ ADB + Area of trapezium BDEC – Area of ∆ AEC

Exercise 8.2 Class 12 Maths Ncert Solutions Question 5.
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.
Solution:
The equations of sides of the triangle are
y = 2x+ 1 … (1), y = 3x + 1 … (2), x = 4 … (3)
Solving equations (1) and (2), we get x = 0 and y = 1
Let A be the point (0,1)
Solving equations (1) and (3), we get x = 4 and y = 9.
Let B be the point (4, 9) Solving equations (2) and (3) we get x = 4 and y = 13
Let C be the point (4, 13)

Ex 8.2 Class12 NCERT Solutions Question 6.
Smaller area bounded by the circle x² + y² = 4 and the line x + y = 2
(a) 2 (π – 2)
(b) π – 2
(c) 2π – 1
(d) 2(π + 2)
Solution:
x² + y² = 4 intersects the positive x-axis at (2, 0) and the positive y-axis at (0, 2). The line x + y = 2 passes through (2, 0) and (0, 2).

Required area is shaded in the figure. Required Area = (Area of circle – Area under the line) lying in the first quadrant

Ex 8.2 Class 12 Ncert Solutions Question 7.
Area lying between the curves y² = 4x and y = 2x.
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 1 }{ 4 }\)
(d) \(\frac { 3 }{ 4 }\)
Solution:
(b) \(\frac { 1 }{ 3 }\)

The given curves are y² = 4x … (1)
y = 2x … (2)
4x(x – 1) = 0 ⇒ x = 0, x = 1
When x = 0, y = 0
When x = 1, y = 2
The points of intersection are (0, 0) and (1, 2)
Required area is shaded in the figure.

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1/

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1

Question 1.
Find the area of the region bounded by the curve y² = x and the lines x = 1, x = 4, and the x-axis.
Solution:
The required area is shaded in the figure and lies between the lines x = 1 and x = 4. Hence the limits of integration are 1 and 4.

Question 2.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and x-axis in the first quadrant
Solution:
The required area is shaded in the figure lies between the lines x = 2 and x = 4. Hence the limits of integration are 2 and 4.

Question 3.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Solution:
x² = 4y ∴ x = 2\(\sqrt{y}\)
The required area is shaded in the figure and lies between the lines y = 2 and y = 4. Hence the limits of integration are 2 and 4.

Question 4.
Find the area of the region bounded by the ellipse \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1
Solution:
i. Put y = o in \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1, we get \(\frac{x^{2}}{16}=1\)
i.e., x² = 16 ∴ x = ± 4
∴ The points of intersection with x-axis are (-4, 0) and (4, 0)

ii. The shaded area lies between x = 0 and x = 4.
∴ The limits of integration are 0 and 4.
The curve is symmetrical w.r.t both the axes
∴ Required area = 4 x area of the shaded region
\(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 }\) = 1 i.e., \(\frac{y^{2}}{9}=1\) – \(\frac{x^{2}}{16}=1\)

Question 5.
Find the area of the region bounded by the ellipse \(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } \) = 1
Solution:

The points of intersection of the curve and the x-axis is obtained by substituting y = 0 in the equation of the curve.
We get \(\frac{x^{2}}{4}=1\), x² = 4 ∴ x = ±2
The curve intersects the x-axis at (-2, 0) and (2,0).
\(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 9 } \) = 1 ∴ \(\frac{y^{2}}{9}\) = 1 – \(\frac{x^{2}}{4}=1\)
= \(\frac{4-x^{2}}{4}\)
∴ y² = \(\frac { 9 }{ 4 }\)(4 – x²) ∴ y = \(\frac{3}{2} \sqrt{4-x^{2}}\)
The curve is symmetrical about both the axes.
Area of the ellipse = 4 x area of the shaded region

Question 6.
Find the area of the region in the first quadrant enclosed by x-axis, line x = \(\sqrt{3}\)y and the circle x² + y² = 4.
Solution:

The given equations are x = \(\sqrt{3}\)y … (1)
x² + y = 4 … (2)
To get the point of intersection of the line and the circle, solve (1) and (2)
From (1), we get y = \(\frac{x}{\sqrt{3}}\)
∴ (2) → x² + \(\left(\frac{x}{\sqrt{3}}\right)^{2}\) = 4 ⇒ x² + \(\frac{x^{2}}{3}\) = 4
\(\frac{4x^{2}}{3}\) = 4 ⇒ x² = 3 ⇒ x = ±\(\sqrt{3}\)
To get the point of intersection of the circle and x-axis puty = 0 in (2)
(2) → x² = 4 ∴x = ± 2
∴ The x coordinate of A and B are \(\sqrt{3}\) and 2 respectively.
Required area = Area of OAC + Area of ACB

Question 7.
Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = \(\frac { a }{ \sqrt { 2 } }\).
Solution:

x² + y² = a² is a circle which intersects the positive x axis at (a, 0).
x² + y² = a²
∴ y = \(\sqrt{a^{2}-x^{2}}\)
The required area is shaded in the figure and lies between x = \(\frac { a }{ \sqrt { 2 } }\) and x – a
∴ Limits of integration are \(\frac { a }{ \sqrt { 2 } }\) and a
Area of the shaded region = 2 x Area of the shaded region in the first quadrant

Question 8.
The area between x = y² and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Solution:

The given curves are x = y² … (1)
x = 4 ….. (2) and x = a … (3)
Since the line x = a divides the area in two equal parts, we get \(2 \int_{0}^{a} y d x=2 \int_{a}^{4} y d x\)
Since the areas are symmetric w.r.t. x-axis

Question 9.
Find the area of the region bounded by the parabola y = x² and y = |x|.
Solution:
Clearly x² = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.
y = |x| i.e., y = x and y = – x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.
The required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis.
So,
Required area = 2 (shaded area in the first quadrant)

Question 10.
Find the area bounded by the curve x² = 4y and the line x = 4y – 2
Solution:

The given curves are x² = 4y … (1)
and x – 4y – 2 … (2)
Substituting (2) in (1),
we get (4y – 2)² = 4y
⇒ 16y² – 16y + 4 = 4y
⇒ 16y² – 20y + 4 = 0
⇒ 4y² – 5y + 1 = 0
⇒ 4y² – 4y – y + 1 = 0
⇒ 4y(y – 1) – 1(y – 1) = 0
⇒ (y – 1)(4y – 1) = 0
⇒ y = 1, y = \(\frac { 1 }{ 4 }\)
When y = 1, x = 2
When y = \(\frac { 1 }{ 4 }\), x = – 1
∴ Points of intersection are (-1, \(\frac { 1 }{ 4 }\)) and (2, 1)
The required area is shaded in the figure. Required area = (Area under the line x = 4y – 2) – (Area under the parabola x² = 4y) :

Question 11.
Find the area of the region bounded by the curve y² = 4x and the line x = 3.
Solution:

The given curves are y² = 4x … (1)
and x = 3
The shaded region is symmetric w.r.t. x-axis.
∴ Required area = 2 x Shaded area in the first quadrant

Question 12.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is
(a) π
(b) \(\frac { \pi }{ 2 } \)
(c) \(\frac { \pi }{ 3 } \)
(d) \(\frac { \pi }{ 4 } \)
Solution:
(a) π

Question 13.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is
(a) 2
(b) \(\frac { 9 }{ 4 }\)
(c) \(\frac { 9 }{ 3 }\)
(d) \(\frac { 9 }{ 2 }\)
Solution:

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-7-ex-7-11/

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.11

Ex 7.11 Class 12 NCERT Solutions Question 1.
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 }x\quad dx }\)
Solution:

7.11 Class 12 NCERT Solutions Question 2.
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx } \)
Solution:

Exercise 7.11 Maths Class 12 Question 3.
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\)
Solution:

7.11 Maths Class 12 NCERT Solutions Question 4.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x}\)
Solution:

Exercise 7.11 Class 12  Question 5.
\(\int_{-5}^{5}|x+2| d x\)
Solution:
|x+2| = x + 2 if x + 2 ≥ 0 ⇒ x ≥ – 2
|x+2| = -(x + 2) if x + 2 < 0 ⇒ x < – 2

Ex 7.11 Class 12 Maths Ncert Solutions Question 6.
\(\int_{2}^{8}|x-5| d x\)
Solution:

Ex7.11 Class 12 NCERT Solutions Question 7.
\(\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I\)
Solution:

Class 12 Ex 7.11 NCERT Solutions Question 8.
\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx } \)
Solution:

7.11 Class 12 Maths Question 9.
\(\int_{0}^{2} x \sqrt{2-x} d x\)
Solution:

Class 12 Maths Exercise 7.11 Solutions Question 10.
\(\int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x)\)dx
Solution:

Question 11.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x\)
Solution:
Here f(x) = sin²x
∴ f(x) = sin²(-x) = [sin(-x)]²
= (-sinx)² = sin²x
Since f(-x) = f(x),
f(x) = sin²x is an even function.

Question 12.
\(\int_{0}^{\pi} \frac{x d x}{1+\sin x}\)
Solution:

Question 13.
\(\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx\)
Solution:
Let f(x) = sin⁷ xdx
⇒ f(-x) = -sin⁷ x = -f(x)
⇒ f(x) is an odd function of x
⇒ \(\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx=0\)

Question 14.
\(\int _{ 0 }^{ 2\pi }{ { cos }^{ 5 } } xdx\)
Solution:
Let f(x) = cos⁵ x
f(2π – x) = cos⁵(2π – x) = cos⁵ x
f(2π – x) = f(x)

Question 15.
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx } \)
Solution:

Question 16.
\(\int_{0}^{\pi} \log (1+\cos x) d x\)
Solution:

Question 17.
\(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x\)
Solution:

Question 18.
\(\int_{0}^{4}|x-1| d x\)
Solution:

Question 19.
Show that \(\int_{0}^{a} f(x) g(x) d x=2 \int_{0}^{a} f(x) d x\) if f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4
Solution:

Question 20.
The value of
\(\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx } \) is
(a) 0
(b) 2
(c) π
(d) 1
Solution:
(c) π
Let f(x) = x³ + x cosx + tan⁵x
f(- x) = (- x)³ + (- x)cos(- x) + tan5(- x)
= – x³ – x cosx – tan5x
= – f(x)
∴ f(x) is an odd function

Question 21.
The value of \(\int_{0}^{\frac{x}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x\) is
(a) 2
(b) \(\frac { 3 }{ 4 }\)
(c) 0
(d) – 2
Solution:

I = – 1
2I = 0 Hence I = 0
Evaluate the following.

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-7-ex-7-8/

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.8

Question 1.
\(\int_{a}^{b} x d x\)
Solution:
Let I = \(\int_{a}^{b} x d x\)
f(x) = x, nh = b – a
f(a) = a
f[a + h) = a + h
f(a + 2 h) = a+ 2h,
……………………….
f[a + (n – 1)h) = a + (n – 1)h

Question 2.
\(\int_{0}^{5}(x+1) d x\)
Solution:
Let I = \(\int_{0}^{5}(x+1) d x\)
We have a = 0, b = 5 and f(x) = x + 1
nh = b-a = 5 – 0 = 5
f(0) = 0 + 1 = 1
f(0 + h) = 0 + h + 1 = h + 1
f(0 + 2h) = 0 + 2h + 1 = 2h + 1
………………………….
f(0 + (n – 1 )h) = 0 + (n – 1)h + 1 = (n – 1)h + 1

Question 3.
\(\int_{2}^{3} x^{2} d x\)
Solution:
Let I = \(\int_{2}^{3} x^{2} d x\)
f(x) = x², a = 2, b = 3, nh = b – a = 3 – 2 = 1
f(2) = 2² = 4,
f(2 + h) = (2 + h)² = 4 + h² + 4h
f(2 + 2h) = (2 + 2h)² = 4+ 4h² + 8h
………………………….
f(2 + (n – 1 )h) = (2 + (n – 1)h]² = 4 + (h – 1)²h² + 4(n – 1)h

Question 4.
\(\int_{1}^{4}\left(x^{2}-x\right) d x\)
Solution:
Let I = \(\int_{1}^{4}\left(x^{2}-x\right) d x\)
We have a = 1, b = 4, f(x) = x² – x and nh = b – a = 4 – 1 = 3
f(1) = 1² – 1 = 0
f(1 + h) = (1 + h)² – (1 + h) = h² + h
f(1 + 2h) = (1 + 2h)² – (1 + 2 h) = 2²h² + 2 h
………………………….
f(1 + n – 1)h) = [1 + (n – 1)h]² – [1 + (n – 1)h] = (n – 1)² h² +(n – 1 )h

Question 5.
\(\int_{-1}^{1} e^{x} d x\)
Solution:
Let I = \(\int_{-1}^{1} e^{x} d x\)
We have a = – 1, b = 1, f(x) = e^x, nh = b – a = 1 + 1 = 2
f(- 1) = e^-1
f(- 1 + h) = e^-1+h
f(- 1 + 2h) = e^-1+2h
………………………….
f(- 1 + (n – h)h) = e^-1+(n-1)h

Question 6.
\(\int_{0}^{4}\left(x+e^{2 x}\right) d x\)
Solution:
Let I = \(\int_{0}^{4}\left(x+e^{2 x}\right) d x\)
f(x) = x + e^2x
a = 0, b = 4, nh = b – a = 4 – 0 = 4
f(0) = 0 + e⁰ = 1
f(0 + h) = (0 + h) + e^2(0+h) = h + e^2h
f(0 + 2h) = (0 + 2h) + e^2(0+h) = 2h + e^2h
………………………….
f(0 + (n – 1)h) = (0 + (n – 1 )h) + e^2(0+h)=(n – 1 )h + e^2(n-1)h