Integration JEE Main Questions: Important Questions, Previous Year Questions with Solutions

Integration JEE Questions with Solutions

The Integration JEE Main questions help you prepare effectively for the JEE Main exam. Solving integration JEE Main questions with solutions is one of the best ways to prepare for IIT JEE.

Question 1. Let e be the base of natural logarithm and let f:{1,2,3,4}1,e,e2,e3 and
g:1,e,e2,e31,12,13,14 be two bijective functions such that f is strictly decreasing and g is strictly increasing. If (x)=f-1g-112x, then the area of the region $\mathrm{R}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2 \leq \mathrm{y} \leq \phi(\mathrm{x})\right.$, 0x1} is :

1. 3-e(2)3e(2)

2. 13e(2)

3. 3+e(2)

4. 3+e(2)2+e(3)

Correct Answer: Option (1)

Solution:

$\begin{aligned} & f(x)=e^{x-1} \\ & g(x)=\frac{1}{\log _e x+1} \\ & f^{-1}\left(g^{-1}(1 / 2)\right)=f^{-1}(e)=2 \\ & \phi(x)=2^x \\ & \therefore \int_0^1\left(2^x-x^2\right) d x=\left(\frac{2^x}{\ln 2}-\frac{x^3}{3}\right)_0^1=\frac{1}{\ln 2}-\frac{1}{3} \\ & =\frac{3-\ln 2}{3 \ln 2}\end{aligned}$

Question 2: The area of the region {(x,y):0y6-x, $\left.\mathrm{y}^2 \geq 4 \mathrm{x}-3, \mathrm{x} \geq 0\right\}$ is :

1. 8

2. 9

3. 12

4. 15

Correct Answer: Option (2)

Solution:

A=036-y-y2+34dy=9

Question 3: The value of the integral -44324x1+exdx is :

1. 4+2

2. 3+8

3. 3+4

4. 4+3

Correct Answer: Option (2)

Solution:

Apply (P-5)

I=-/4/43241+e-d

$\begin{aligned}
& \text { Add } 2 \mathrm{I}=\int_{-\pi / 4}^{\pi / 4} 32 \cos ^4 \theta \mathrm{~d} \theta=2 \int_0^{\pi / 4} 32 \cos ^4 \theta \mathrm{~d} \theta \\
& \mathrm{I}=32 \int_0^{\pi / 4} \cos ^4 \theta \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4}\left(2 \cos ^2 \theta\right)^2 \mathrm{~d} \theta=8 \int_0^{\pi / 4}(1+\cos 2 \theta)^2 \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4} 1+2 \cos 2 \theta+\cos ^2 2 \theta \mathrm{~d} \theta \\
& =8 \int_0^{\pi / 4} 1+2 \cos 2 \theta+\frac{1+\cos 4 \theta}{2} \mathrm{~d} \theta \\
& =8\left[\frac{3 \theta}{2}+\frac{2 \sin 2 \theta}{2}+\frac{\sin 4 \theta}{8}\right]_0^{\pi / 4} \\
& =8\left[\left(\frac{3 \pi}{8}+1+0\right)\right]=3 \pi+8
\end{aligned}$

Question 4: The value of the integral -11x3+|x|+1x2+2|x|+1dx is equal to

1. 3e2

2. 2e2

3. 5e3

4. 3e3

Correct Answer: Option (2)

Solution: $\begin{aligned} & I=\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=\int_{-1}^1 \frac{x^3}{x^2+2|x|+1} \cdot d x+\int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} \cdot d x \\ & I=0+\int_{-1}^1 \frac{|x|+1}{|x|^2+2|x|+1} \cdot d x \\ & 2 \int_0^1 \frac{1}{x+1} \cdot d x \\ & I=2[\ln |x+1|]_0^1 \\ & I=2 \ln (2)\end{aligned}$

Question 5: The area of the region (x,y):x2-8xy-x is:

1. 3436

2. 6376

3. 4376

4. 5236

Correct Answer: Option (1)

Solution: $\begin{aligned} & \text { Required area }=\int_0^7\left(7 x-x^2\right) d x \\ & =\left[\frac{7 x^2}{2}-\frac{x^3}{3}\right]_0^7=\frac{343}{6}\end{aligned}$

Question 6: If /6/4x-3x+3+1dx=e(3-1), then 92 is equal to ____ .

1. 12

2. 13

3. 11

4. 14

Correct Answer: Option (1)

Solution:

$\begin{aligned} & \int_{\pi / 6}^{\pi / 4}\left(\left(\cot \left(\mathrm{x}-\frac{\pi}{3}\right) \cot \left(\mathrm{x}+\frac{\pi}{3}\right)+1\right) \mathrm{dx}\right. \\ & \text { using } \cot (\mathrm{A}-\mathrm{B})=\frac{\cot \mathrm{B} \cot \mathrm{A}+1}{\cot \mathrm{~B}-\cot \mathrm{A}} \\ & \Rightarrow \int_{\pi / 6}^{\pi / 4}-\frac{1}{\sqrt{3}}\left[\cot \left(\mathrm{x}-\frac{\pi}{3}\right)-\cot \left(\mathrm{x}+\frac{\pi}{3}\right)\right] \mathrm{dx} \\ & \Rightarrow \frac{-1}{\sqrt{3}}\left[\ln \left|\frac{\sin \left(\mathrm{x}-\frac{\pi}{3}\right)}{\sin \left(\mathrm{x}+\frac{\pi}{3}\right)}\right|\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\end{aligned}$

$\begin{aligned} & =\frac{-1}{\sqrt{3}}\left[\ln \left|\left(\frac{\sin 15^{\circ}}{\left.\sin 105^{\circ}\right)}\right)\right|-\ell \mathrm{n}\left|\frac{\sin 30^{\circ}}{\sin 90^{\circ}}\right|\right] \\ & =\frac{-1}{\sqrt{3}}\left[\ln \left(\tan 15^{\circ}\right)-\ell \mathrm{n}\left(\frac{1}{2}\right)\right. \\ & =\frac{-1}{\sqrt{3}}[\ln (2-\sqrt{3})+\ell \mathrm{n} 2] \\ & \Rightarrow \frac{-1}{\sqrt{3}} \ln (4-2 \sqrt{3})=\frac{-2}{\sqrt{3}} \ln (\sqrt{3}-1) \\ & \therefore \alpha=\frac{-2}{\sqrt{3}} \therefore 9 \alpha^2=12\end{aligned}$

Question 7: Let $f(x)=\left\{\begin{array}{lll}\frac{1}{3} & , & x \leq \pi / 2 \\ \frac{b(1-\sin x)}{(\pi-2 x)^2} & , & x>\pi / 2\end{array}\right.$, If f is continuous at x=/2, then the value of 03~b-6x2+2x-3dx is :

1. 5

2. 2

3. 3

4. 4

Correct Answer: Option (4)

Solution:

x2b1-2-x42-x2=13b=83

02x2+2x-3dx=02|(x+3)(x-1)|dx

put x-1=t

-11|(t+4)t|dt=-10-t2-4tdt+01t2+4tdt

-t33-2t2-10+t33+2t201=4

Question 8: The value of the integral
02xx2+x+1(x+1)x4+x2+1dx is equal to :

1. 13e(3-22)

2. 23e(4+2)

3. 23e(3+22)

4. 13e(1+62)

Correct Answer: Option (3)

Solution: $\begin{aligned} & I=\int_0^2 \sqrt{\frac{x\left(x^2+x+1\right)}{(x+1)\left(x^2-x+1\right)\left(x^2+x+1\right)} d x} \\ & =\int_0^2 \sqrt{\frac{x}{x^3+1}} d x \\ & \left(x^{3 / 2}=t \Rightarrow \frac{3}{2} x^{1 / 2} d x=d t\right) \\ & =\frac{2}{3} \int_0^{2^{3 / 2}} \frac{d t}{\sqrt{t^2+1}} \\ & =\frac{2}{3}\left(\ln \left(t+\sqrt{t^2+1}\right)\right)_0^{2^{3 / 2}} \\ & =\frac{2}{3} \ln \left(2^{3 / 2}+3\right)\end{aligned}$

Question 9: The value of the integral 0e(x)x2+4dx is:

1. e(2)2

2. e(2)4

3. 1+e(2)

4. 2+e(2)

Correct Answer: Option (2)

Solution:

Put x=2tdx=2dt

$\begin{aligned}
& \mathrm{I}=\int_0^{\infty} \frac{\ell \mathrm{n} 2 \mathrm{t}}{4 \mathrm{t}^2+4}(2 \mathrm{dt})=\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{n} 2+\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \\
& =\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{n} 2}{\mathrm{t}^2+1} \mathrm{dt}+\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \\
& {\left[\frac{\ln 2}{2} \mathrm{tan}^{-1} \mathrm{t}\right]_0^{\infty}+\mathrm{I}_1} \\
& =\frac{\ell \mathrm{n} 2}{2} \cdot \frac{\pi}{2}+\mathrm{I}_1 \\
& \mathrm{I}_2=\frac{1}{2} \int_0^{\infty} \frac{\ell \mathrm{nt}}{\mathrm{t}^2+1} \mathrm{dt} \quad \mathrm{t}=\frac{1}{\mathrm{u}} \Rightarrow \mathrm{dt}=-\frac{\mathrm{du}}{\mathrm{u}^2} \\
& =\frac{1}{2} \int_{\infty}^0 \frac{\ell \mathrm{n}(1 / \mathrm{u})}{\frac{1}{\mathrm{u}^2}+1}\left(-\frac{\mathrm{du}}{\mathrm{u}^2}\right)
\end{aligned}$

Add I1=0

$\mathrm{I}=\frac{\pi \ell \mathrm{n} 2}{4}$

Question 10: The integral 01-11+x+x2dx is equal to :

1. 2-12+12e54+2

2. 2-12+12e54-2

3. 2-12-12e54+2

4. 2-12-12e54-2

Correct Answer: Option (4)

Solution: $\begin{aligned}= & \int_0^1 \tan ^{-1}\left(\frac{1}{1+x+x^2}\right) d x \\ = & \int_0^1 \tan ^{-1}\left(\frac{(x+1)-x}{1+x(x+1)}\right) d x \\ = & \int_0^1\left(\tan ^{-1}(x+1)-\tan ^{-1} x\right) d x \\ = & \left(x \tan ^{-1}(x+1)-\frac{1}{2} \ell \mathrm{n}\left|1+(1+x)^2\right|+\tan ^{-1}(x+1)\right)_0^1 \\ & -\left(x \tan ^{-1} x-\frac{1}{2} \ell \mathrm{n}\left|1+x^2\right|\right)_0^1 \\ = & \left(2 \tan ^{-1} 2-\frac{\pi}{4}-\frac{1}{2} \ell \mathrm{n} \frac{5}{2}\right)-\left(\frac{\pi}{4}-\frac{1}{2} \ell \mathrm{n} 2\right) \\ = & 2 \tan ^{-1} 2-\frac{1}{2} \ell \mathrm{n}\left(\frac{5}{4}\right)-\frac{\pi}{2}\end{aligned}$

Question 11: Let -22(|x|+[xx])dx=2(3-2)+, where [•] is the greatest integer function. Then 2 equals.

1. 1

2. 2

3. 4

4. 8

Correct Answer: Option (2)

Solution: $\begin{aligned} & \int_{-2}^2(|\sin x|+[x \sin x]) d x \\ & \Rightarrow 2 \int_0^2(|\sin x|+[x \sin x]) d x \\ & \Rightarrow 2\left[\int_0^2 \sin x d x+\int_0^2[x \sin x] d x\right] \\ & 2\left([-\cos x]_0^2+\int_0^\alpha 0 d x+\int_\alpha^2 2 d x\right) \\ & 2(3-\cos 2-2-\alpha) \\ & 2(3-\cos 2)-2 \alpha \\ & \text { where } \alpha \text { is Root of } x \sin x=1 \\ & \text { Now } \operatorname{comparing} \beta=-2 \alpha \\ & \alpha=-\frac{\beta}{2} \\ & -\frac{\beta}{2} \sin \left(\frac{-\beta}{2}\right)=1 \\ & \beta \sin \left(\frac{\beta}{2}\right)=2\end{aligned}$

Question 12: The value of 0204x+4xdx is equal to:

1. 152

2. 25

3. 15

4. 252

Correct Answer: Option (3)

Solution: $\begin{aligned} & I=\int_0^{20 \pi}\left(1-2 \sin ^2 x \cos ^2 x\right) \cdot d x \\ & =20 \pi-\frac{1}{2} \int_0^{20 \pi} \sin ^2 2 x d x \\ & 20 \pi-\frac{1}{2}(40) \int_0^{\frac{\pi}{2}} \sin ^2 2 x \\ & 20 \pi-\frac{20}{2}\left(x-\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{2}} \\ & 20 \pi-10\left(\frac{\pi}{2}\right)=15 \pi\end{aligned}$

Question 13: Let f(x)=dxx23+2x12 be such that f(0)=-26+24e(2). If f(1)=a+be(3), where a,bz, then a+b is equal to:

1. -18

2. -11

3. -5

4. -26

Correct Answer: Option (2)

Solution:

f(x)=dxx2/3+2x1/2

Put x=t6dx=6t5dt

$\begin{aligned}
& =\int \frac{6 t^5 d t}{t^4+2 t^3}=6 \int \frac{\left(t^2-4\right)+4}{t+2} d t \\
& =6\left[\int(t-2) d t+4 \int \frac{1}{t+2} d t\right] \\
& =6\left[\frac{\mathrm{t}^2}{2}-2 \mathrm{t}+4 \ln (\mathrm{t}+2)\right]+\mathrm{C} \\
& =3 \mathrm{x}^{1 / 3}-12 \mathrm{x}^{1 / 6}+24\left(\mathrm{n}\left(\mathrm{x}^{16}+2\right)+\mathrm{C}\right. \\
& \mathrm{f}(0)=24(\mathrm{n} 2+\mathrm{C}=-26+24(\mathrm{n} 2 \text { (given) } \\
& \Rightarrow \mathrm{C}=-26
\end{aligned}$

Now

$\begin{aligned}
& f(1)=-35+24(n 3=A+B(n 3 \text { (as given in ques.) } \\
& \Rightarrow A=-35 \& B=24 \\
& \Rightarrow A+B=-11
\end{aligned}$

Hence, the correct answer is option (2).

Question 14: The value of r=1200rx|x|dx is ____

1. 210

2. 110

3. 112

4. 230

Correct Answer: Option (1)

Solution:

$\begin{aligned} & I_r=\int_0^r x|\sin \pi x| d x \\ & \text { King's Rule }=\int_0^r(r-x)|\sin \pi x| d x \\ & 2 I_r=\int_0^r r|\sin \pi x| d x \\ & I_r=\frac{r}{2} \int_0^r|\sin \pi x| d x \\ & I_1=\frac{1}{2} \int_0^1|\sin \pi x| d x=\frac{1}{2 \pi} \int_0^\pi|\sin t| d t=\frac{1}{2 \pi} \cdot 2 \\ & I_2=\frac{2}{2} \int_0^2|\sin \pi x| d x=\frac{2}{2 \pi} \int_0^{2 \pi}|\sin t| d t=\frac{2}{2 \pi} .4\end{aligned}$

$\begin{aligned} & \therefore S=\sqrt{\pi \cdot \frac{1}{2 \pi} \cdot 2}+\sqrt{\pi \cdot \frac{2}{2 \pi} \cdot 4}+\sqrt{\pi \cdot \frac{3}{2 \pi} \cdot 6 \ldots \ldots \ldots .+\sqrt{\pi \cdot \frac{20}{2 \pi} \cdot 2(20)}} \\ & =1+2+3+\ldots \ldots .20=\frac{20 \times 21}{2}=210\end{aligned}$

Hence, the correct answer is 210.

Question 15: If 1-52x5x2xdx=f(x)+C, where C is the constant of integration, then f6-f4 is equal to

1. 13(26-3)

2. 23(4+6)

3. 43(8-6)

4. 13(26+3)

Correct Answer: Option (3)

Solution:

$\begin{aligned} & \int \frac{1-5 \cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\ & \int \frac{\sec ^2 x d x}{\sin ^5 x}-\int \frac{5}{\sin ^5 x} d x \\ & \int \frac{1}{\sin ^5 x} \tan x-\int \frac{-5}{\sin ^6 x} \cos x \cdot \tan x d x-\int \frac{5}{\sin ^6 x} d x=\frac{\tan x}{\sin ^5 x}=f(x) \\ & f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)=\frac{32}{\sqrt{3}}-\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^5}=\frac{32}{\sqrt{3}}-4 \sqrt{2}=\frac{4}{\sqrt{3}}(8-\sqrt{6})\end{aligned}$

Hence, the correct answer is option 3.

Question 16: If 1-52x5x2xdx=f(x)+C, where C is the constant of integration, then f6-f4 is equal to

1. 13(26-3)

2. 23(4+6)

3. 43(8-6)

4. 13(26+3)

Correct Answer: Option (3)

Solution: $\begin{aligned} & \int \frac{1-5 \cos ^2 x}{\sin ^2 x \cos ^2 x} d x \\ & \int \frac{\sec ^2 x d x}{\sin ^5 x}-\int \frac{5}{\sin ^5 x} d x \\ & \int \frac{1}{\sin ^5 x} \tan x-\int \frac{-5}{\sin ^6 x} \cos x \cdot \tan x d x-\int \frac{5}{\sin ^6 x} d x=\frac{\tan x}{\sin ^5 x}=f(x) \\ & f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)=\frac{32}{\sqrt{3}}-\frac{1}{\left(\frac{1}{\sqrt{2}}\right)^5}=\frac{32}{\sqrt{3}}-4 \sqrt{2}=\frac{4}{\sqrt{3}}(8-\sqrt{6})\end{aligned}$

Hence, the correct answer is option 3.

Question 17: Let $f(t)=\int\left(\frac{\left(1-\sin \left(\log _e t\right)\right.}{1-\cos \left(\log _e t\right)}\right) d t, t>1$. if. fe/2=-e/2 and fe/4=e/4 then equals:

1. -1-22

2. -1+2

3. -1-2

4. 1+2

Correct Answer: Option (3)

Solution:

Let t=xt=exdt=exdx

$\begin{aligned} & G . I=\int \frac{1-\sin x}{2 \sin ^2 \frac{x}{2}} e^x d x=\frac{1}{2} \int\left(\cos e^2 \frac{x}{2}-2 \cot \frac{x}{2}\right) e^x d x=\frac{1}{2} e^x\left(-2 \cot \frac{x}{2}\right)+c \\ &\Rightarrow-t \cot \left(\frac{\operatorname{los} t}{2}\right)+c \\ & f\left(e^{-\pi / 2}\right)=-e^{\pi / 2} \quad \Rightarrow c=0 \\ & \therefore d\left(e^{\pi / 4}\right)=-e^{-\pi / 4}(\sqrt{2}+1) \\ & \therefore \alpha=-(\sqrt{2}+1)\end{aligned}$

Hence, the correct answer is option (3).

Question 18: Let I(x)=3dx(4x+6)4x2+8x+3 and I(0)=34+20.
If I12=a2b+c, where $a, b, c, \in N, \operatorname{gcd}(a, b)=1$, then a+b+c is equal to:

1. 26

2. 23

3. 25

4. 24

Correct Answer: Option (1)

Solution:

I(x)=32dx(2x+3)(2x+3)(2x+1)=32dx(2x+3)3/2(2x+1)1/2

$\begin{aligned} & =\frac{3}{2} \int \frac{d x}{\left(\frac{2 x+3}{2 x+1}\right)^{3 / 2}(2 x+1)^2}\left(t=\frac{2 x+3}{2 x+1} \Rightarrow d t=\frac{-4}{(2 x+1)^2} d x\right) \\ & =\frac{3}{2} \int\left(\frac{2 x+3}{2 x+1}\right)^{-\frac{3}{2}} \frac{d x}{(2 x+1)^2}=\frac{-3}{8} \frac{\left(\frac{2 x+3}{2 x+1}\right)^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+C=\frac{3}{4}\left(\frac{2 x+1}{2 x+3}\right)^{\frac{1}{2}}+C \\ & I(0)=20+\frac{\sqrt{3}}{4} \Rightarrow c=20 \\ & I\left(\frac{1}{2}\right)=\frac{3 \sqrt{2}}{8}+20 \\ & a+b+c=3+8+20=26\end{aligned}$

Hence the correct answer is option (1).

Question 19: The value of the integral 24524~dx1+32x is:

1. 18

2. 6

3. 3

4. 12

Correct Answer: Option (4)

Solution:

Let the given integral be I :

$
I=\int_{\frac{\pi}{2 \pi}}^{\frac{5 \pi}{3 \pi}} \frac{d x}{1+\sqrt{\tan 2 x}}
$
The sum of the limits is a+b=24+524=624=4.
Using the property abf(x)dx=abf(a+b-x)dx, replace x with 4-x :

$
\begin{aligned}
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\sin }{2 \pi}} \frac{d x}{1+\sqrt[3]{\tan 2\left(\frac{\pi}{4}-x\right)}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\sin }{2 \pi}} \frac{d x}{1+\sqrt[3]{\tan \left(\frac{\pi}{2}-2 x\right)}}
\end{aligned}
$
Since 2-= :

$
\begin{aligned}
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\frac{\pi}{2 \pi}}{2 \pi}} \frac{d x}{1+\sqrt[3]{\cot 2 x}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\pi}{2 \pi}} \frac{d x}{1+\frac{1}{\sqrt[3]{\tan 2 x}}} \\
& I=\int_{\frac{\pi}{2 \pi}}^{\frac{\pi}{2 \pi}} \frac{\sqrt[3]{\tan 2 x}}{1+\sqrt{\tan 2 x}} d x
\end{aligned}
$
Now, add the two expressions for I :

$
\begin{aligned}
& 2 I=\int \frac{\frac{\pi}{\pi}}{\frac{\pi}{2 \pi}}\left(\frac{1}{1+\sqrt[3]{\tan 2 x}}+\frac{\sqrt[3]{\tan 2 x}}{1+\sqrt[3]{\tan 2 x}}\right) d x \\
& 2 I=\int \frac{\frac{\pi}{2}}{\frac{1+\sqrt[3]{\tan 2 x}}{1+\sqrt[3]{\tan 2 x}} d x} \\
& 2 I=\int \frac{\frac{\pi}{2 \pi}}{\frac{\pi}{24}} 1 d x \\
& 2 I=[x] \frac{\frac{\pi x}{\pi}}{24} \\
& 2 I=\frac{5 \pi}{24}-\frac{\pi}{24}=\frac{4 \pi}{24}=\frac{\pi}{6} \\
& I=\frac{\pi}{12}
\end{aligned}
$

Hence, the correct answer is option (4).

Question 20: The value of -221[x]+4dx, where [.] denotes the greatest integer function, is:

1. 160(-7)

2. 760(3-1)

3. 760(-3)

4. 160(21-1)

Correct Answer: Option (2)

Solution:

I=-2-11-2+4dx+-101-1+4dx+0110+4dx+1211+4dx
$ I=\frac{1}{2}\left(-1+\frac{\pi}{2}\right)+\frac{1}{3}(0-(-1))+\frac{1}{4}(1-0)+\frac{1}{5}\left(\frac{\pi}{2}-1\right)$
$ I=\frac{\pi}{4}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{\pi}{10}-\frac{1}{5} $
$ I=\frac{7 \pi}{20}-\frac{7}{60}$
I=760(3-1)

Hence, the correct answer is option (2).

Question 21: If 014-11-2x+4x2dx=a-1(2)-be(5), where abN, then (2a+b) is equal to...

1. 9

2. 5

3. 6

4. 4

Correct Answer: Option (1)

Solution:

$\begin{aligned}
& 4 \int_0^1 \operatorname{Tan}^{-1}\left(\frac{1}{1-2 x+4 x^2}\right) d x \\
& 4 \int_0^1 \operatorname{Tan}^{-1}\left(\frac{2 x-(2 x-1)}{1+2 x+(2 x-1)}\right) d x \\
& I=4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x-4 \int_0^1 \operatorname{Tan}^{-1}(2 x-1) d x \ldots \ldots .(i) \\
& 4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x-4 \int_0^1 \operatorname{Tan}^{-1}(2(1-x)-1) d x \\
& I=4 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x+4 \int_0^1 \operatorname{Tan}^{-1}(2 x-1) d x \ldots . .(i i) \\
& \text { (i) }+ \text { (ii) } 2 I=8 \int_0^1 \operatorname{Tan}^{-1} 2 x . d x
\end{aligned}$

By integration by parts $I=4 \operatorname{Tan}^1 2-\log 5$

$\therefore 2 a+b=8+1=9$

Hence, the correct answer is 9.

Question 22: 60|(3x+2x+x)|dx is equal to:

1. 17

2. 21

3. 24

4. 26

Correct Answer: Option (1)

Solution:

$\begin{aligned}
& \text { Sol: } 6 \int_0^\pi|\sin 3 x+\sin 2 x+\sin x| d x \\
& =6 \int_0^\pi|\sin 2 x+2 \sin 2 x \cos x| d x=6 \int_0^\pi|\sin 2 x(1+2 \cos x)| d x \\
& =6 \int_0^\pi|2 \cos x(1+2 \cos x)| \sin x d x \quad \cos x=t \\
& =12 \int_1^{-1}|t(1+2 t)|-d t \quad-\sin x \cdot d x=d t
\end{aligned}$

$\begin{aligned}
& =12 \int_{-1}^1|t(1+2 t)| d t=12\left[\int_{-1}^{-1 / 2}\left(t+2 t^2\right) d t-\int_{-1 / 2}^0\left(t+2 t^2\right) d t+\int_0^1\left(t+2 t^2\right) d t\right] \\
= & 12\left[\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)_{-1}^{-1 / 2}-\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)_{-1 / 2}+\left(\frac{t^2}{2}+\frac{2 t^3}{3}\right)\right] \\
= & 12\left[\left(\frac{1}{8}-\frac{1}{12}\right)-\left(\frac{1}{2}-\frac{2}{3}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+\left(\frac{1}{2}+\frac{2}{3}\right)\right] \\
= & 12\left[\frac{2}{8}-\frac{2}{12}+\frac{4}{3}\right]=12\left[\frac{1}{4}-\frac{1}{6}+\frac{4}{3}\right]=3-2+16=17
\end{aligned}$

OR,

$\begin{aligned}
& 6 \int_0^\pi|2 \sin 2 x \cos x+\sin 2 x| d x \\
& 6 \int_0^\pi\left|4 \sin x \cos ^2 x+2 \sin \cos x\right| d x
\end{aligned}$

$I=12 \int_0^\pi \sin x\left|2 \cos ^2 x+\cos x\right| d x \quad \text { Put } \cos x=t,-\sin x d x=d t$

Hence, the correct answer is 17.

Question 23: If a function f(x) satisfies a condition f(x)=x+0/2xyf(y) dy, then f4 is equal to:

1. 1322+2-

2. 4-32

3. 2-22

4. 126-4-

Correct Answer: Option (4)

Solution:

Let, f(x)=x+kx
where k=0/2yf(y)dy

$
\begin{aligned}
& \Rightarrow k=\int_0^{\pi / 2}\left(\sin y \cdot \cos y+k \sin ^2 y\right) d y=-\frac{\cos 2 y}{4}+\left.\frac{k}{2}\left(y-\frac{\sin 2 y}{2}\right)\right|_0 ^{\pi / 2} \\
& \Rightarrow k=\frac{1}{2}+\frac{k \pi}{4} \Rightarrow k=\frac{2}{4-\pi}
\end{aligned}
$

Hence f(x)=x+24-x

Question 24: Evaluate x2x+x-1dx

1. -121+2x2+c

2. 121+3x2+c

3. -121+2 x2+c

4. 121+3x2+c

Correct Answer: Option (3)

Solution:

$\begin{aligned}
& I=\int \frac{\sec x d x}{2 \tan x+\sec x-1} \\
& =\int \frac{d x}{2 \sin x+1-\cos x}=\int \frac{d x}{4 \sin \frac{x}{2} \cos \frac{x}{2}+2 \sin ^2 \frac{x}{2}} \\
& =\frac{1}{2} \int \frac{\operatorname{cosec}^2 \frac{x}{2} d x}{2 \cot \frac{x}{2}+1}
\end{aligned}$

Let $2 \cot \frac{\mathrm{x}}{2}=\mathrm{t} \Rightarrow-\operatorname{cosec}^2 \frac{\mathrm{x}}{2} \mathrm{dx}=\mathrm{dt}$
Hence I=-12dtt+1=-12|1+t|+ constant

$=-\frac{1}{2} \ln \left|1+2 \cot \frac{x}{2}\right|+\text { c }$

Hence, the correct answer is option (3).

Question 25: 40113+x2+1+x2dx-3e(3) is equal to:

1. 2+2+e(1+2)

2. 2-2-e(1+2)

3. 2+2-e(1+2)

4. 2-2+e(1+2)

Correct Answer: Option (2)

Solution:

Given the expression:

40113+x2+1+x2dx-3e(3)

First, rationalize the denominator inside the integral by multiplying numerator and denominator by

3+x2-1+x2

This gives:

013+x2-1+x2(3+x2)2-(1+x2)2dx

Since

(3+x2)2-(1+x2)2=(3+x2)-(1+x2)=2

the integral becomes

013+x2-1+x22dx=12013+x2-1+x2dx

Therefore, the original expression is

412013+x2-1+x2dx-3e(3)=2013+x2-1+x2dx-32e3

Split the integral:

2013+x2dx-011+x2dx-32e3

Recall the integral formula

x2+cdx=x2x2+c+c2x+x2+c+C

Apply for c=3:

013+x2dx=x2x2+3+32x+x2+301

Evaluate at limits:

=124+32(1+2)-0+323=1+323-343=1+343

Similarly for c=1:

011+x2dx=x2x2+1+12x+x2+101

Evaluate at limits:

=122+12(1+2)-0=22+12(1+2)

Substitute back:

21+343-22-12(1+2)-323

Simplify:

=2-2+323-(1+2)-323=2-2-(1+2)

Hence, the value of the original expression is

2-2-(1+2)=2-2-e(1+2)

Hence, the correct answer is option (2).

Integration JEE Last 10 Years' Most Asked Topics

JEE Main Integration questions over the last 10 years have covered various topics. The table below shows the most asked JEE Main and JEE Advanced integration question topics.

Also Check: JEE Main 2027 Important Formulas

Important Integration Formulas for JEE Main

The important formulas to solve Integration JEE Main questions are provided below; students must learn them:

1. Power Rule of Integration
   xn,dx=xn+1n+1+C,n-1

2. Logarithmic Integration Formula
   1x,dx=|x|+C

3. Exponential Function Formula
   ex,dx=ex+C

4. General Exponential Formula
   ax,dx=axa+C

5. Integration of Sine Function
   x,dx=-x+C

6. Integration of Cosine Function
   x,dx=x+C

7. Integration of Secant Square
   2x,dx=x+C

8. Standard Inverse Tangent Formula
   11+x2,dx=-1x+C

9. Standard Inverse Sine Formula
   11-x2,dx=-1x+C

10. Integration by Parts Formula
    u,dv=uv-v,du

Best Books for Integration JEE Main Questions

While preparing for IIT JEE, it is very important to follow the best books because they help properly cover all the concepts. Refer to the books given below for solving Integration JEE Main questions and answers.

Related JEE Main Maths Preparation Resources