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.

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JEE Main Definite Integration Past Year Questions With Solutions

Question 1: If n is any integer, then find the value of 0πecos2xcos3(2n+1)xdx\int_{0}^{\pi }{{{e}^{{{\cos }^{2}}x}}{{\cos }^{3}}(2n+1)x\,dx}

Solution:

Since cos(2n+1)(πx)=cos[(2n+1)π(2n+1)x]=cos(2n+1)x and cos2(πx)\cos (2n+1)(\pi -x)=\cos [(2n+1)\pi -(2n+1)x] = -\cos (2n+1)x \text \ and \ {{\cos }^{2}}(\pi -x) =cos2x={{\cos }^{2}}x

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

0πecos2xcos3(2n+1)xdx=0.\int_{0}^{\pi }{{{e}^{{{\cos }^{2}}x}}{{\cos }^{3}}(2n+1)x\,dx=0}.

Question 2: Let f be a positive function. Let I1=1kkxf{x(1x)}dx{{I}_{1}}=\int_{1-k}^{k}{x\,f\left\{ x(1-x) \right\}}\,dx

 

I2=1kkf{x(1x)}dx{{I}_{2}}=\int_{1-k}^{k}{\,f\left\{ x(1-x) \right\}}\,dx when 2k − 1 > 0. Then find I1/I2{{I}_{1}}/{{I}_{2}}.

Solution:

I1=1kkxf{x(1x)}dx=1kk(1k+kx)f[(1k+kx){1(1k+kx)}]dx{{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 abf(x)dx=abf(a+bx)dx)=1kk(1x)f{x(1x)}dx=1kkf{x(1x)}dx1kkxf{x(1x)}dx=I2I1 Therefore, 2I1=I2I1I2=12 \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 01(1+cos8x)(ax2+bx+c)dx=02(1+cos8x)(ax2+bx+c)dx\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 ax2+bx+c=0a{{x}^{2}}+bx+c=0 has ____ root(s).

Solution:

We have 02f(x)dx=01f(x)dx+12f(x)dx,\int_{0}^{2}{f(x)dx=\int_{0}^{1}{f(x)dx+\int_{1}^{2}{f(x)dx}}}, where f(x)=(ax2+bx+c)(1+cos8x)f(x)=(a{{x}^{2}}+bx+c)(1+{{\cos }^{8}}x)

If f(x)>0(<0)x(1,2)f(x)>0(<0)\,x\in (1,\,2) then 12f(x)dx>0(<0).\int_{1}^{2}{f(x)dx>0(<0)}.

Thus f(x)=(1+cos8x)(ax2+bx+c)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+cos8x)1(1+{{\cos }^{8}}x)\ge 1 follows that if g(x)=ax2+bx+c,g(x)=a{{x}^{2}}+bx+c, then there exist some α,β(1,2)\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 ax2+bx+c=0a{{x}^{2}}+bx+c=0 has at least one root in (1, 2) and hence in (0, 2).

Question 4: The value of ππcos2x1+axdx,a>0,\int_{\,-\,\pi }^{\,\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}dx,\,a>0,} is

Solution:

I=ππcos2x1+axdx=ππcos2x1+ax(dx)=ππcos2x1+axdxI+I=ππcos2x(11+ax+11+ax)dx=ππcos2xdx2I=20πcos2x.dx=0π(1+cos2x)dx2I=[x]0π+[sin2x2]0π2I=πI=π2.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)=01tm(1+t)ndt,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)=01tm(1+t)ndt[(1+t)ntm+1m+1]0101n(1+t)n1tm+1m+1dt=2nm+1nm+1l(m+1,n1).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×(12×1×1)=2=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 π/23π/2[2sinx]dx\int\limits_{\pi /2}^{3\pi /2}{[2\sin x]\,dx} is

Solution:

We know

1sinx122sinx2I=π23π2[2sinx]dx=π25π6[2sinx]dx+5π6π[2sinx]dx+π7π6[2sinx]dx+7π63π2[2sinx]dx=π25π6(1)dx+5π6π(0)dx+π7π6(1)dx+7π63π2(2)dx=(5π6π2)+0(7π6π)2(3π27π6)=2π6π64π6=π2.-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)=Asin(πx2)+B,f(12)=2f(x)=A\sin \left( \frac{\pi x}{2} \right)+B, {f}’\left( \frac{1}{2} \right)=\sqrt{2} and 01f(x)dx=2Aπ,\int_{0}^{1}{f(x)\,dx=\frac{2A}{\pi },} then find the constants A and B respectively.

Solution:

f(x)=Asin(πx2)+B,f(12)=201f(x)dx=2Aπ,01{Asin(πx2)+B}dx=2Aπ2Aπcosπx2+Bx01=2AπB(2Aπ)=2AπB=0 therefore f(x)=Asinπx2f(x)=πA2cosπx2 and f(12)=πA2(12)=2πA=4A=4πA=4π,B=0.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 0πsin(n+12) xsinxdx,(nN)\int\limits_{0}^{\pi }{\,\frac{\sin \left( n+\frac{1}{2} \right)\text{ }x}{\sin x}}\,dx, (n\in N)

Solution:

2sinx2.(12+cosx+cos2x+..+cosnx)=sinx2+2sinx2cosx+2sinx2cos2x+.+2sinx2cosnx=sinx2+sin3x2sinx2+sin5x2sin3x2+..+sin(n+12)xsin(n12)x=sin(n+12)x12+cosx+cos2x+..+cosnx=sin(n+12)x2sin(x2)0πsin(n+12)xsin(x2)dx=2(0π12dx+0πcosxdx+..+0πcosnxdx)=2(π2+sinx+..+sinnxn)0π=π.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 π4,π4,5π4.\frac{\pi }{4}, \frac{\pi }{4},\frac{5\pi }{4}.

Since, sinxcosx\sin x\ge \cos x on the interval [π4,5π4]\left[ \frac{\pi }{4},\frac{5\pi }{4} \right]

Therefore, area of one such region =π/45π/4(sinxcosx) dx=22 sq. unit.=\int_{\pi /4}^{5\pi /4}{(\sin x-\cos x)\ dx}\\ =2\sqrt{2} \text \ sq. \ unit.

Question 11: The integral π12π48cos2x(tanx+cotx)3\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

Answer: (1)

Solution:

Let I = π12π48cos2x(tanx+cotx)3\int_{\frac{\pi}{12}}^{\frac{\pi}{4}}\frac{8cos2x}{(tanx + cotx)^3} dx

JEE Past Year Questions on Definite Integration

I=π12π48  cos2x×sin3(2x)8  dxI = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}}{8 \; cos2x} \times \frac{sin^3(2x)}{8} \; dx

 

I=π12π4(sin2x  cos2x)×sin2(2x)    dxI = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} (sin2x \; cos2x) \times sin^2(2x) \; \; dx

 

I=12π12π4sin4x  (1cos4x2)    dxI = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} sin4x \; (\frac{1-cos4x}{2})\; \; dx

 

I=14π12π4sin4x  dx18π12π4sin8x  dxI = \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

I = 15/128 (Answer!)

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

JEE Past Year Questions on Definite Integrals

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

Adding equation (4) to given, we have

Definite Integrals Past Year Solved Questions

Question 13: Find the value of 018  log(1+x)1+x2  dx\int_0^1 \frac{8 \; log(1+x)}{1+x^2} \; dx

Solution:

018  log(1+x)1+x2  dx\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 = 0π48log(1+tanθ)dθ\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

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