Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 17 of 23 · Click an option or "Show Solution" to reveal answer

Q161
Let f,g:(0,∞)→Rf, g:(0, \infty) \rightarrow \mathbb{R} be two functions defined by f(x)=∫−xx(∣t∣−t2)e−t2dtf(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d t and g(x)=∫0x2t1/2e−tdtg(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d t. Then, the value of 9(f(log⁡e9)+g(log⁡e9))9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right) is equal to :
A 10
B 9
C 8
D 6
Correct Answer
Option C
Solution
f(x)=∫−xx(∣t∣−t2)e−t2dt⇒f′(x)=2⋅(∣x∣−x2)e−x2………..(1)g(x)=∫0x2t12e−tdtg′(x)=xe−x2(2x)−0f′(x)+g′(x)=2xe−x2−2x2e−x2+2x2e−x2\begin{aligned} & \mathrm{f}(\mathrm{x})=\int\limits_{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\ & \mathrm{g}(\mathrm{x})=\int\limits_0^{\mathrm{x}^2} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-t} \mathrm{dt} \\ & \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}^2}(2 \mathrm{x})-0 \\ & \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{xe}^{-\mathrm{x}^2}-2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}+2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2} \end{aligned}

Integrating both sides w.r.t.

x\mathrm{x}
f(x)+g(x)=∫0α2xe−x2dxx2=t⇒∫0αe−tdt=[−e−t]0α=−e(log⁡c(9)−1)+1⇒9(f(x)+g(x))=(1−19)9=8\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\int\limits_0^\alpha 2 \mathrm{xe}^{-\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{x}^2=\mathrm{t} \\ & \Rightarrow \int_0^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_0^{\sqrt{\alpha}} \\ & =-\mathrm{e}^{\left(\log _c(9)^{-1}\right)+1} \\ & \Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8 \end{aligned}
Q162
lim⁡x→π2(1(x−π2)2∫x3(π2)3cos⁡(t13)dt)\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right) is equal to
A 3π24\dfrac{3 \pi^2}{4}
B 3π28\dfrac{3 \pi^2}{8}
C 3π4\dfrac{3 \pi}{4}
D 3π8\dfrac{3 \pi}{8}
Correct Answer
Option B
Solution

Using L'hospital rule

=lim⁡x→π−20−cos⁡x×3x22(x−π2)=lim⁡x→π−2sin⁡(x−π2)2(x−π2)×3π24=3π28\begin{aligned} & =\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim \limits_{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\ & =\frac{3 \pi^2}{8} \end{aligned}
Q163
If the value of the integral ∫−π2π2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡x2123)dx=π4(π+a)−2\int\limits_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\left(\dfrac{x^2 \cos x}{1+\pi^x}+\dfrac{1+\sin ^2 x}{1+e^{\sin x^{2123}}}\right) d x=\dfrac{\pi}{4}(\pi+a)-2, then the value of aa is
A −32-\dfrac{3}{2}
B 3
C 32\dfrac{3}{2}
D 2
Correct Answer
Option B
Solution
I=∫−π/2π/2(x2cos⁡x1+πx+1+sin⁡2x1+esin⁡x2023)dxI=∫−π/2π/2(x2cos⁡x1+π−x+1+sin⁡2x1+esin⁡(−x)2023)dx\begin{aligned} & I=\int\limits_{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x \\ & I=\int\limits_{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x \end{aligned}

On Adding, we get

2I=∫−π/2π/2(x2cos⁡x+1+sin⁡2x)dx2 I=\int\limits_{-\pi / 2}^{\pi / 2}\left(x^2 \cos x+1+\sin ^2 x\right) d x

On solving

I=π24+3π4−2a=3\begin{aligned} & \mathrm{I}=\frac{\pi^2}{4}+\frac{3 \pi}{4}-2 \\ & \mathrm{a}=3 \end{aligned}
Q164
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function defined by f(x)=x(1+x4)1/4f(x)=\dfrac{x}{\left(1+x^4\right)^{1 / 4}}, and g(x)=f(f(f(f(x))))g(x)=f(f(f(f(x)))). Then, 18∫025x2g(x)dx18 \int_0^{\sqrt{2 \sqrt{5}}} x^2 g(x) d x is equal to
A 36
B 33
C 39
D 42
Correct Answer
Option C
Solution
f(x)=x(1+x4)1/4f∘f(x)=f(x)(1+f(x)4)1/4=x(1+x4)1/4(1+x41+x4)1/4=x(1+2x4)1/4f(f(f(f(x))))=x(1+4x4)1/4\begin{aligned} &f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\ &f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\ &f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}} \end{aligned}
18∫025x3(1+4x4)1/4dx18 \int\limits_0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x
 Let 1+4x4=t416x3dx=4t3dt184∫13t3dtt=92(t33)13=32[26]=39\begin{aligned} & \text{ Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\ & 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\ & \frac{18}{4} \int\limits_1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\ & =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\ & =\frac{3}{2}[26]=39 \end{aligned}
Q165
Let y=f(x)y=f(x) be a thrice differentiable function in (−5,5)(-5,5). Let the tangents to the curve y=f(x)y=f(x) at (1,f(1))(1, f(1)) and (3,f(3))(3, f(3)) make angles π/6\pi / 6 and π/4\pi / 4, respectively with positive xx-axis. If 27∫13((f′(t))2+1)f′′(t)dt=α+β327 \int\limits_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} where α,β\alpha, \beta are integers, then the value of α+β\alpha+\beta equals
A 26
B −-16
C 36
D −-14
Correct Answer
Option A
Solution
y=f(x)⇒dydx=f′(x)dydx)(1,f(1))=f′(1)=tan⁡π6=13⇒f′(1)=13dydx)(3,f(3))=f′(3)=tan⁡π4=1⇒f′(3)=127∫13((f′(t))2+1)f′′(t)dt=α+β3I=∫13((f′(t))2+1)f′′(t)dtf′(t)=z⇒f′′(t)dt=dzz=f′(3)=1z=f′(1)=13\begin{aligned} & y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\ & \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \\ & \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 \\ & 27 \int\limits_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \\ & I=\int\limits_1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t \\ & f^{\prime}(t)=z \Rightarrow f^{\prime \prime}(t) d t=d z \\ & z=f^{\prime}(3)=1 \\ & z=f^{\prime}(1)=\frac{1}{\sqrt{3}} \end{aligned}
I=∫1/31(z2+1)dz=(z33+z)1/31=(13+1)−(13⋅133+13)=43−1093=43−10273α+β3=27(43−10273)=36−103α=36,β=−10α+β=36−10=26\begin{aligned} & I=\int\limits_{1 / \sqrt{3}}^1\left(z^2+1\right) d z=\left(\frac{z^3}{3}+z\right)_{1 / \sqrt{3}}^1 \\ & =\left(\frac{1}{3}+1\right)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}\right) \\ & =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} \\ & \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}\right)=36-10 \sqrt{3} \\ & \alpha=36, \beta=-10 \\ & \alpha+\beta=36-10=26 \end{aligned}
Q166
Let aa and bb be real constants such that the function ff defined by f(x)={x2+3x+a,x≤1bx+2,x>1f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right. be differentiable on R\mathbb{R}. Then, the value of ∫−22f(x)dx\int\limits_{-2}^2 f(x) d x equals
A 21
B 19/6
C 17
D 15/6
Correct Answer
Option C
Solution

To determine the integral of the piecewise function

ff

over the interval

[−2,2][-2, 2]

, we first ensure that

ff

is differentiable on

R\mathbb{R}

, as given in the problem statement. Differentiability implies continuity, so

ff

must also be continuous at

x=1x=1

. The condition for continuity at

x=1x=1

is:

x2+3x+a=bx+2x^2 + 3x + a = bx + 2

at

x=1x=1

. This simplifies to:

1+3+a=b(1)+21 + 3 + a = b(1) + 2
⇒a=b−2\Rightarrow a = b - 2

The first derivative of

ff

gives us two different expressions depending on the value of

xx

: For

x≤1x \leq 1

,

f′(x)=2x+3f'(x) = 2x + 3

; and for

x>1x > 1

,

f′(x)=bf'(x) = b

. For

ff

to be differentiable at

x=1x=1

, these derivatives must be equal at that point. Setting

f′(1)f'(1)

from both expressions equal to each other gives

2(1)+3=b2(1) + 3 = b

, thus

b=5b = 5

. And from

a=b−2a = b - 2

, we have

a=3a = 3

. Now, we can calculate the integral of

ff

over the specified interval:

∫−21(x2+3x+3) dx+∫12(5x+2) dx\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx
=[x33+3x22+3x]−21+[5x22+2x]12=(13+32+3)−(−83+6−6)+(10+4−52−2)=6+32+12−52=17\begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\ & =6+\frac{3}{2}+12-\frac{5}{2}=17 \end{aligned}
Q167
The value of lim⁡n→∞∑k=1nn3(n2+k2)(n2+3k2)\lim \limits_{n \rightarrow \infty} \sum\limits_{k=1}^n \dfrac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)} is :
A π8(23+3)\dfrac{\pi}{8(2 \sqrt{3}+3)}
B (23+3)π24\dfrac{(2 \sqrt{3}+3) \pi}{24}
C 13π8(43+3)\dfrac{13 \pi}{8(4 \sqrt{3}+3)}
D 13(23−3)π8\dfrac{13(2 \sqrt{3}-3) \pi}{8}
Correct Answer
Option C
Solution
lim⁡n→∞∑k=1nn3n4(1+k2n2)(1+3k2n2)=lim⁡n→∞1n∑k=1nn3(1+k2n2)(1+3k2n2)\begin{aligned} & \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \end{aligned}
=∫01dx3(1+x2)(13+x2)=\int\limits_0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}
=∫0113×32(x2+1)−(x2+13)(1+x2)(x2+13)dx=12∫01[1x2+(13)2−11+x2]dx=12[3tan⁡−1(3x)]01−12(tan⁡−1x)01=32(π3)−12(π4)=π23−π8=13π8⋅(43+3)\begin{aligned} & =\int\limits_0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\ & =\frac{1}{2} \int\limits_0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\ & =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\ & =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\ & =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)} \end{aligned}
Q168
Let f:[−π2,π2]→Rf:\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \rightarrow \mathbf{R} be a differentiable function such that f(0)=12f(0)=\dfrac{1}{2}. If the lim⁡x→0x∫0xf(t)dtex2−1=α\lim \limits_{x \rightarrow 0} \dfrac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alpha, then 8α28 \alpha^2 is equal to :
A 4
B 2
C 1
D 16
Correct Answer
Option B
Solution
lim⁡x→0x∫0xf(t)dt(ex2−1x2)×x2lim⁡x→0∫0xf(t)dtx(lim⁡x→0ex2−1x2=1)=lim⁡x→0f(x)1 (using L Hospital) f(0)=12α=128α2=2\begin{aligned} & \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\ & \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\ & =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text{ (using L Hospital) } \\\\ & f(0)=\frac{1}{2} \\\\ & \alpha=\frac{1}{2} \\\\ & 8 \alpha^2=2 \end{aligned}
Q169
The integral ∫1/43/4cos⁡(2cot⁡−11−x1+x)dx\int\limits_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\dfrac{1-x}{1+x}}\right) d x is equal to
A −1/2-1/2
B −1/4-1/4
C 1/4
D 1/2
Correct Answer
Option B
Solution
∫1434cos⁡(2cot⁡−11−x1+x)dxx=cos⁡2θ⇒dx=(−2sin⁡2θdθ)\begin{aligned} & \int\limits_{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\ & x=\cos 2 \theta \\ & \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta) \end{aligned}

Take limit as α\alpha and β\beta

−2∫αβcos⁡2θ⋅sin⁡2θdθ=∫αβsin⁡4θdθ=−cos⁡4θ4∣αβ=−14(2⋅(x2)−1)∣1/43/4=−14(2x2−1)∣1/43/4=−14(1816−1−216+1)=−14\begin{aligned} & -2 \int\limits_\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\ & =\int\limits_\alpha^\beta \sin 4 \theta d \theta \\ & =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\ & =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\ & =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\ & =-\frac{1}{4} \end{aligned}
Q170
lim⁡x→π2(∫x3(π/2)3(sin⁡(2t1/3)+cos⁡(t1/3))dt(x−π2)2)\lim \limits_{x \rightarrow \dfrac{\pi}{2}}\left(\dfrac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\dfrac{\pi}{2}\right)^2}\right) is equal to
A 3π22\dfrac{3 \pi^2}{2}
B 9π28\dfrac{9 \pi^2}{8}
C 5π29\dfrac{5 \pi^2}{9}
D 11π210\dfrac{11 \pi^2}{10}
Correct Answer
Option B
Solution
lim⁡x→π2(∫x3(π/2)3(sin⁡(2t1/3)+cos⁡(t1/3))dt(x−π2)2)\lim \limits_{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)

Using Newton Leibniz theorem

=lim⁡x→π2[sin⁡(2×π2)⋅0−sin⁡(2x)⋅3x2+(cos⁡π2)⋅0−cos⁡x⋅3x22(x−π2)]=lim⁡wx→π2−3x2sin⁡2x−3x2cos⁡x2(x−π2)(00) form =lim⁡x→π2[−6xsin⁡2x−6x2cos⁡2x−6xcos⁡x+3x2sin⁡x2]=6×π24+3×π242=9π28\begin{aligned} & =\lim \limits_{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\ & =\lim \limits_ w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text{ form } \\ & =\lim \limits_{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\ & =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\ & =\frac{9 \pi^2}{8} \end{aligned}
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