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
\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]
\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]
\[\int\limits_0^\infty e^{- x} dx .\]
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
