Advertisements
Advertisements
प्रश्न
Evaluate the integral by using substitution.
`int_1^2 (1/x- 1/(2x^2))e^(2x) dx`
Advertisements
उत्तर
Let `I = int_1^2 e^(2x) (1/x - 1/(2x^2)) dx`
Put 2x = t
⇒ 2dx = dt
When x = 1, t = 2
And when x = 2, t = 4
∴ `I = 1/2 int_2^4 e^t (2/t - (1 xx4)/(2t^2)) dt`
`= 1/2 int_2^4 e^t (2/t - 2/t^2) dt`
`= int_2^4 e^t* (1/t - 1/t^2) dt`
`= int_2^4 e^t *[1/t + d/dt (1/t)] dt`
`= [e^t * 1/t]_2^4 = 1/4 e^4 - e^2/2`
`= e^2/2 (e^2/2 - 1)`
or `(e^2 (e^2 - 2))/4`
APPEARS IN
संबंधित प्रश्न
Evaluate `int_(-1)^2|x^3-x|dx`
Evaluate :
`∫_0^π(4x sin x)/(1+cos^2 x) dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate the integral by using substitution.
`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)
Evaluate the integral by using substitution.
`int_(-1)^1 dx/(x^2 + 2x + 5)`
`int 1/(1 + cos x)` dx = _____
A) `tan(x/2) + c`
B) `2 tan (x/2) + c`
C) -`cot (x/2) + c`
D) -2 `cot (x/2)` + c
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate:
Evaluate the following integral:
\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate
\[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\]
Evaluate the following integral:
Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .
Evaluate: `int_-1^2 (|"x"|)/"x"d"x"`.
If `I_n = int_0^(pi/4) tan^n theta "d"theta " then " I_8 + I_6` equals ______.
`int_0^1 x(1 - x)^5 "dx" =` ______.
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^(pi4) sec^4x "d"x` = ______.
For the integral \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which substitution is suitable?
After using \[t=\tan^{-1}x\] in \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which integral results?
