Advertisements
Advertisements
Question
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Sum
Solution
`int_0^1 x"e"^(x^2) "d"x = 1/2 int_0^1 2x"e"^(x^2) "d"x`
Let t = x2
Then dt = 2x dx
When x = 0, t = 0
x = 1, t = 1
So the integral becomes,
`1/2int_0^2 "e"^"t" "dt" = 1/2 ["e"^"t"]_0^1`
= `1/2 ["e" - 1]`
shaalaa.com
Definite Integrals
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\]
\[\int\limits_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx\]
\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} dx\]
\[\int\limits_0^2 e^x dx\]
\[\int\limits_{- 1}^1 x\left| x \right| dx .\]
\[\int\limits_0^{\pi/2} x \sin x\ dx\] is equal to
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
Find `int sqrt(10 - 4x + 4x^2) "d"x`
