# JEE Main Maths Definite Integrals Previous Year Questions With Solutions

JEE Previous year questions on the topic definite integrals are available here. Before we start solving problems, let us have a brief look on, what is definite integrals? “A definite integral is an integral with lower and upper limits”. This article covers definite integral questions from the past year of JEE Main along with the detailed solution for each question. These questions include all the important topics and formulae. Students can also increase their problem-solving efficiency by referring to the solved examples and practising them. About 2-4 questions are asked from this topic in JEE Examination every year. Aspirant can download free pdf and practice all the questions and get ready for the exam.

## JEE Main Definite Integration Past Year Questions With Solutions

Question 1: If n is any integer, then find the value of $\int_{0}^{\pi }{{{e}^{{{\cos }^{2}}x}}{{\cos }^{3}}(2n+1)x\,dx}$

Solution:

Since $\cos (2n+1)(\pi -x)=\cos [(2n+1)\pi -(2n+1)x] = -\cos (2n+1)x \text \ and \ {{\cos }^{2}}(\pi -x)$ $={{\cos }^{2}}x$

So that f (2a − x) = −f (x), and hence by the property of definite integral

$\int_{0}^{\pi }{{{e}^{{{\cos }^{2}}x}}{{\cos }^{3}}(2n+1)x\,dx=0}.$

Question 2: Let f be a positive function. Let ${{I}_{1}}=\int_{1-k}^{k}{x\,f\left\{ x(1-x) \right\}}\,dx$

${{I}_{2}}=\int_{1-k}^{k}{\,f\left\{ x(1-x) \right\}}\,dx$ when 2k − 1 > 0. Then find ${{I}_{1}}/{{I}_{2}}$.

Solution:

${{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}dx}\\ =\int_{1-k}^{k}{(1-k+k-x)f[(1-k+k-x)\{1-(1-k+k-x)\}]dx}$

(Because $\int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)dx)}}\\ =\int_{1-k}^{k}{\,\,(1-x)f\{x(1-x)\}}\,dx\\ =\int_{1-k}^{k}{f\{x(1-x)\}}\,dx-\int_{1-k}^{k}{xf\{x(1-x)\}}\,dx={{I}_{2}}-{{I}_{1}}\\ \text \ Therefore, \ 2{{I}_{1}}={{I}_{2}}\Rightarrow \frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2}$

Question 3: Let a, b, c be non-zero real numbers such that $\int_{0}^{1}{(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)\,dx}=\int_{0}^{2}{(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)\,dx}$ Then the quadratic equation $a{{x}^{2}}+bx+c=0$ has ____ root(s).

Solution:

We have $\int_{0}^{2}{f(x)dx=\int_{0}^{1}{f(x)dx+\int_{1}^{2}{f(x)dx}}},$ where $f(x)=(a{{x}^{2}}+bx+c)(1+{{\cos }^{8}}x)$

If $f(x)>0(<0)\,x\in (1,\,2)$ then $\int_{1}^{2}{f(x)dx>0(<0)}.$

Thus $f(x)=(1+{{\cos }^{8}}x)(a{{x}^{2}}+bx+c)$ must be positive for some value of x in [1, 2] and must be negative for some value of x in [1, 2].

As $(1+{{\cos }^{8}}x)\ge 1$ follows that if $g(x)=a{{x}^{2}}+bx+c,$ then there exist some $\alpha ,\beta \in (1,\,2)$ such that g (α) > 0 and g (β) < 0. Since g is continuous on R, therefore there exists some c between αand βsuch that g (c) = 0.

Thus $a{{x}^{2}}+bx+c=0$ has at least one root in (1, 2) and hence in (0, 2).

Question 4: The value of $\int_{\,-\,\pi }^{\,\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}dx,\,a>0,}$ is

Solution:

$I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}dx}=\int_{\,\pi }^{\,-\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}(-dx)} \\ =\int_{\,-\pi }^{\,\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}}\,dx\\ \Rightarrow I+I=\int_{\,-\pi }^{\,\pi }{{{\cos }^{2}}x\left( \frac{1}{1+{{a}^{x}}}+\frac{1}{1+{{a}^{-x}}} \right)\,dx}\\ =\int_{\,-\pi }^{\,\pi }{{{\cos }^{2}}x\,dx}\\ 2I =2\int_{0}^{\pi }{{{\cos }^{2}}x.\,dx=}\int_{0}^{\pi }{(1+\cos 2x)dx}\\ 2I =[x]_{0}^{\pi }+\left[ \frac{\sin 2x}{2} \right]_{0}^{\pi }\\ \Rightarrow 2I=\pi \,\,\,\Rightarrow I=\frac{\pi }{2}.$

Question 5: If $l(m,\,n)=\int_{0}^{1}{{{t}^{m}}{{(1+t)}^{n}}dt,}$ then the expression for l (m, n) in terms of l (m + 1, n − 1) is

Solution:

$l(m,n)=\int_{0}^{1}{{{t}^{m}}(1+t}{{)}^{n}}dt\\ \left[ {{(1+t)}^{n}}\frac{{{t}^{m+1}}}{m+1} \right]_{0}^{1}-\int_{0}^{1}{n{{(1+t)}^{n-1}}\frac{{{t}^{m+1}}}{m+1}}\,dt\\ =\frac{{{2}^{n}}}{m+1}-\frac{n}{m+1}l(m+1,n-1).$

Question 6: The area bounded by the curves y = |x| − 1 and y = −|x| + 1 is

Solution:

The lines are y = x − 1 , x ≥ 0

y = −x − 1, x < 0, y = −x + 1, x ≥ 0, y = x + 1, x < 0

Area $=4\times \left( \frac{1}{2}\times 1\times 1 \right)=2$

Question 7: If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral $\int\limits_{\pi /2}^{3\pi /2}{[2\sin x]\,dx}$ is

Solution:

We know

$-1\le \sin x\le 1\Rightarrow -2\le 2\sin x\le 2\\ I=\int_{\frac{\pi }{2}}^{\frac{3\pi }{2}}{[2\sin x]dx}\\ =\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}{[2\sin x]dx+\int_{\frac{5\pi }{6}}^{\pi }{[2\sin x]dx}}+\int_{\pi }^{\frac{7\pi }{6}}{[2\sin x]dx+\int_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}{[2\sin x]dx}}\\ =\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}{(1)dx+\int_{\frac{5\pi }{6}}^{\pi }{(0)dx+\int_{\pi }^{\frac{7\pi }{6}}{(-1)dx+\int_{\frac{7\pi }{6}}^{\frac{3\pi }{2}}{(-2)}}dx}} \\ =\left( \frac{5\pi }{6}-\frac{\pi }{2} \right)+0-\left( \frac{7\pi }{6}-\pi \right)-2\left( \frac{3\pi }{2}-\frac{7\pi }{6} \right) \\ =\frac{2\pi }{6}-\frac{\pi }{6}-\frac{4\pi }{6}=-\frac{\pi }{2}.$

Question 8: If $f(x)=A\sin \left( \frac{\pi x}{2} \right)+B, {f}’\left( \frac{1}{2} \right)=\sqrt{2}$ and $\int_{0}^{1}{f(x)\,dx=\frac{2A}{\pi },}$ then find the constants A and B respectively.

Solution:

$f(x)=A\sin \left( \frac{\pi x}{2} \right)+B,\,f’\left( \frac{1}{2} \right)=\sqrt{2}\\ \int_{0}^{1}{f(x)dx=\frac{2A}{\pi }}, \\ \int_{0}^{1}{\left\{ A\sin \left( \frac{\pi x}{2} \right)+B \right\}dx=\frac{2A}{\pi }}\\ \left| -\frac{2A}{\pi }\cos \frac{\pi x}{2}+Bx \right|_{0}^{1}=\frac{2A}{\pi }\\ B-\left( \frac{-2A}{\pi } \right)=\frac{2A}{\pi }\Rightarrow B=0\\ \text \ therefore \ \,\,f(x)=A\sin \frac{\pi x}{2}\Rightarrow f'(x)=\frac{\pi A}{2}\cos \frac{\pi x}{2}\\ \text \ and \ \,f’\left( \frac{1}{2} \right)=\frac{\pi A}{2}\left( \frac{1}{\sqrt{2}} \right)=\sqrt{2}\Rightarrow \pi A=4\Rightarrow A=\frac{4}{\pi } \\ A=\frac{4}{\pi },B=0.$

Question 9: Evaluate $\int\limits_{0}^{\pi }{\,\frac{\sin \left( n+\frac{1}{2} \right)\text{ }x}{\sin x}}\,dx, (n\in N)$

Solution:

$2\sin \frac{x}{2}.\left( \frac{1}{2}+\cos x+\cos 2x+…..+\cos nx \right)\\ =\sin \frac{x}{2}+2\sin \frac{x}{2}\cos x+2\sin \frac{x}{2}\cos 2x+….+2\sin \frac{x}{2}\cos nx\\ =\sin \frac{x}{2}+\sin \frac{3x}{2}-\sin \frac{x}{2}+\sin \frac{5x}{2}-\sin \frac{3x}{2}+….. +\sin \left( n+\frac{1}{2} \right)x-\sin \left( n-\frac{1}{2} \right)x=\sin \left( n+\frac{1}{2} \right)x \\ \frac{1}{2}+\cos x+\cos 2x+…..+\cos nx=\frac{\sin \left( n+\frac{1}{2} \right)x}{2\sin \left( \frac{x}{2} \right)}\\ \int_{0}^{\pi }{\frac{\sin \left( n+\frac{1}{2} \right)x}{\sin \left( \frac{x}{2} \right)}dx}=2\left( \int_{0}^{\pi }{\frac{1}{2}dx+\int_{0}^{\pi }{\cos xdx+…..+\int_{0}^{\pi }{\cos nx\,dx}}} \right)\\ =2\left( \frac{\pi }{2}+\sin x+…..+\frac{\sin nx}{n} \right)_{0}^{\pi }=\pi .$

Question 10: The sine and cosine curves intersect infinitely many times giving bounded regions of equal areas. Then find the area of one such region.

Solution:

Point of intersection of y = sin x and y = cos x is $\frac{\pi }{4}, \frac{\pi }{4},\frac{5\pi }{4}.$

Since, $\sin x\ge \cos x$ on the interval $\left[ \frac{\pi }{4},\frac{5\pi }{4} \right]$

Therefore, area of one such region $=\int_{\pi /4}^{5\pi /4}{(\sin x-\cos x)\ dx}\\ =2\sqrt{2} \text \ sq. \ unit.$

Question 11: The integral $\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{8cos2x}{(tanx + cotx)^3}$ dx equals

(1) 15/18           (2) 13/32                (3) 13/256               (4) 15/64

Solution:

Let I = $\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{8cos2x}{(tanx + cotx)^3}$ dx

$I = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}}{8 \; cos2x} \times \frac{sin^3(2x)}{8} \; dx$

$I = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} (sin2x \; cos2x) \times sin^2(2x) \; \; dx$

$I = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} sin4x \; (\frac{1-cos4x}{2})\; \; dx$

$I = \frac{1}{4} \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} sin4x \; dx – \frac{1}{8} \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} sin8x \; dx$

On solving above integral and putting lower and upper limits, we have

Question 12: If f(a + b + 1 – x) = f(x) ∀ x, where a and b are fixed positive real numbers, then below expression is equal to

Solution:

f(a + b + 1 – x) = f(x) …(1)

At x -> x + 1

f(a + b – x) = f(x+1) …(2)

From (1) and (2)

I = $\frac{1}{a+b} \int_a^b (a + b – x)[f(x+1) + f(x)]\; \; dx$ …(4)

Adding equation (4) to given, we have

Question 13: Find the value of $\int_0^1 \frac{8 \; log(1+x)}{1+x^2} \; dx$

Solution:

$\int_0^1 \frac{8 \; log(1+x)}{1+x^2} \; dx$

Put x = tanθ, So θ = tan-1 x

dθ = 1/(1+x2) dx

Now,

Let I = $\int_0^{\frac{\pi}{4}} 8 log(1 + tan \theta)d\theta$  ……(1)

Lets say, θ = π/4 – θ because x + y = π/4

=> (1 + tan x)(1 + tan y) = 2

(1)=> I = π log 2