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
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int_0^1 x\log\left( 1 + 2x \right)dx\]
\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]
Evaluate each of the following integral:
\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]
\[\int\limits_2^3 \frac{1}{x}dx\]
\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Find `int sqrt(10 - 4x + 4x^2) "d"x`
