NIMCET 2027 Definite Integration Previous Year Questions (PYQs) – Page 2 of 4
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Showing 35 questions
If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to
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Solution
Which of the following is TRUE?
A. If $f$ is continuous on $[a,b]$, then $\int ^b_axf(x)\mathrm{d}x=x\int ^b_af(x)\mathrm{d}x$
B. $\int ^3_0{e}^{{x}^2}dx=\int ^5_0e^{{x}^2}dx+{\int ^5_3e}^{{x}^2}dx$
C. If $f$ is continuous on $[a,b]$, then $\frac{d}{\mathrm{d}x}\Bigg{(}\int ^b_af(x)dx\Bigg{)}=f(x)$
D. Both (a) and (b)
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Solution
If for non-zero x, $cf(x)+df\Bigg{(}\frac{1}{x}\Bigg{)}=|\log |x||+3,$ where $c\ne 0$, then $\int ^e_1f(x)dx=$
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Solution
The value of
$\displaystyle \int_0^{\pi/2} \frac{dx}{1+\tan^3 x}$
is
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Solution
Using property
$\int_0^{\pi/2} f(\tan x)dx = \int_0^{\pi/2} f(\cot x),dx$
Add both:
$I=\int_0^{\pi/2} \frac{1}{1+\tan^3 x}dx$
$I=\int_0^{\pi/2} \frac{\tan^3 x}{1+\tan^3 x}dx$
$2I=\int_0^{\pi/2}1dx=\frac{\pi}{2}$
$I=\frac{\pi}{4}$
If $f:\mathbb R\to\mathbb R$ and $g:\mathbb R\to\mathbb R$ are continuous functions, then evaluate
$\displaystyle \int_{-\pi/2}^{\pi/2}[f(x)+f(-x)][g(x)-g(-x)],dx$
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Solution
$f(x)+f(-x)$ is an even function
$g(x)-g(-x)$ is an odd function
Product of even and odd function is odd
Integral of odd function over symmetric limits is $0$
Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ |x - 2| & \text{if } x \geq 0 \end{cases}$. Find $\int_{-2}^{3} f(x)\, dx$
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Solution
The value of $\frac{d}{dx}\int ^{2\sin x}_{\sin {x}^2}{e}^{{t}^2}dt$ at $x=\pi$
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Solution
We want to find
$\dfrac{d}{dx}\left(\int_{\sin^2 x}^{2\sin x} e^{t^2}, dt\right)$ at $x=\pi$.
Using Leibniz rule:
$\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t) dt\right)
= f(b(x)) b'(x) ;-; f(a(x)) a'(x)$
Here
$a(x)=\sin^2 x$,
$b(x)=2\sin x$,
$f(t)=e^{t^2}$.
Compute derivatives:
$a'(x)=2\sin x\cos x$
$b'(x)=2\cos x$
So the derivative is:
$e^{(2\sin x)^2}(2\cos x)-e^{(\sin^2 x)^2}(2\sin x\cos x)$
Now evaluate at $x=\pi$
$\sin\pi=0,\quad \cos\pi=-1$
Thus:
$b'(\pi)=2(-1)=-2$
$a'(\pi)=0$
Therefore:
$e^{0}(-2)-e^{0}(0)=-2$
So the final answer is:
$-2$
The value of the integral
$\displaystyle \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}, dx$ is
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Solution
Use substitution: x → 9 – x
I = ∫ √x / (√(9–x) + √x) dx
I = ∫ √(9–x) / (√x + √(9–x)) dx
Add both expressions:
2I = ∫₃⁶ 1 dx = 3
⇒ I = 3/2
The value of the integral
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + \sin 2x}, dx$ is
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Solution
sin2x = 2 sinx cosx
Rewrite numerator:
sinx + cosx = √2 sin(x + π/4)
Denominator:
3 + sin2x = 3 + 2 sinx cosx = 1 + (sinx + cosx)²
Let t = sinx + cosx
dt/dx = cosx – sinx → use identity to convert dx
Integral evaluates to (1/4) log 3
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and More.
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