CUET UG Mathematics Applied Definite Integration Previous Year Questions (PYQs)
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$\displaystyle \int_{0}^{1}\frac{a-bx^{2}}{\left(a+bx^{2}\right)^{2}}dx\ \text{is:}$
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Solution
The value of the integral $\displaystyle \int_{\log_{e}2}^{\log_{e}3}\frac{e^{2x}-1}{e^{2x}+1},dx$ is:
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Solution
$\displaystyle \int_{0}^{\pi/2}\frac{1-\cot x}{cosec x+\cos x},dx=$
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Solution
$\int_{-1}^{1}(|x-2|+|x|),dx=$
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Solution
On interval $[-1,1]$, we have $x-2<0$
So,
$|x-2|=2-x$
Now split $|x|$ at $x=0$
For $-1\le x\le 0$
$|x|=-x$
Integrand:
$(2-x-x)=2-2x$
For $0\le x\le 1$
$|x|=x$
Integrand:
$(2-x+x)=2$
Now integrate:
$\int_{-1}^{0}(2-2x),dx=[2x-x^2]_{-1}^{0}$
$=0-(-3)=3$
Next,
$\int_{0}^{1}2,dx=2$
Total value:
$3+2=5$
The order of the differential equation whose general solution is
$y = e^x (a\cos x + b\sin x)$, where $a$ and $b$ are arbitrary constants is:
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Solution
Given solution:
$y = e^x (a\cos x + b\sin x)$
Here the solution contains **two arbitrary constants** $a$ and $b$.
The order of a differential equation is equal to the number of arbitrary constants in its general solution.
Therefore,
Order $=2$
$\int \tan x(\sec x - \tan x)\,dx =$
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Solution
$\int \tan x(\sec x - \tan x)dx$
Expand the expression
$=\int (\sec x\tan x - \tan^2 x)dx$
Split the integral
$=\int \sec x\tan x\,dx - \int \tan^2 x\,dx$
We know
$\frac{d}{dx}(\sec x)=\sec x\tan x$
So
$\int \sec x\tan x\,dx = \sec x$
Also
$\tan^2 x = \sec^2 x -1$
Therefore
$\int \tan^2 x\,dx
= \int (\sec^2 x -1)dx
= \tan x - x$
Hence
$\int \tan x(\sec x-\tan x)dx
= \sec x - (\tan x - x)$
$= \sec x - \tan x + x + C$
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
Online Test Series, Information About Examination,
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and More.
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