Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
बेरीज
उत्तर
= `int_1^"e" (1 + logx)^-3/x "d"x`
= `[("f"(x)^(-3 + 1))/(-3 + 1)]_1^"e"`
= `[(1 + log x)^-2/-2]_1^"e"`
= `- 1/2 [[1 + log x]^-2]_1^"e"`
= `- 1/2 [(1 + log "e")^-2 (1 + log 1)^-2]`
= - 1/2 [(1 + 1)^-2 - (1)^-2]`
= `- 1/2 [1/(2)^2 - 1/(1)^2]`
= `- 1/2[1/4 - 1]`
= `-1/2[(1 - 4)/4]`
= `- 1/2[(-3)/4]`
= `3/8`
shaalaa.com
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]
\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]
\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to
\[\int\limits_0^4 x dx\]
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
