Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^1 x\ e^{x^2} d\ x . \]
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = 0, t = 0\ and\ x\ = 1\, t = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 e^t\ dt\]
\[ \Rightarrow I = \frac{1}{2} \left( e^t \right)_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( e - 1 \right)\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
