Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
बेरीज
Advertisements
उत्तर
= `int_0^(1/4) sqrt((1 - 4)^(1/2)) "d"x`
= `[(1 - 4x)^(3/2)/((3/2)(-4))]_0^(1/4)`
= `[(1 - 4x)^(3/2)/(-6)]_0^(1/4)`
= `- 1/6 [(1 - 4x)^(3/2)]_0^(1/4)`
= `- 1/6 [(1 - 4(1/4))^(3/2) - [1 - 4(0)]^(3/2)]`
= `- 1/6 [0 - (1)^(3/2)]`
= `- 1/6 (- 1)`
= `1/6`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]
\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]
\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx\]
\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]
Evaluate the following integral:
\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]
Evaluate each of the following integral:
\[\int_0^\frac{\pi}{4} \sin2xdx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
