Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
योग
Advertisements
उत्तर
`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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]
\[\int\limits_1^2 \log\ x\ dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^3 \left( x + 4 \right) dx\]
\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]
\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
