Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
योग
Advertisements
उत्तर
`int_0^1 "e"^(2x) "d"x = ["e"^(2x)/2]_0^1`
= `1/2 ["e"^(2x)]_0^1`
= `1/2["e"^(2(1)) - "e"^(2(0))]`
= `1/2 ["e"^2 - "e"^0]`
= `1/2 ["e"^2 - 1]`
shaalaa.com
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]
\[\int\limits_0^1 \frac{\tan^{- 1} x}{1 + x^2} dx\]
\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]
\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\] equals to
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
