Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
योग
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]
\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]
\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]
\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
