Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
बेरीज
उत्तर
`int_1^2 (x - 1)/x^2 "d"x = int_1^2 (x/x^2 - 1/x^2) "d"x`
= `int_1^2 (1/x - x^-2) "d"x`
= `int_1^2 1/x "d"x - int_1^2 x^-2 "d"x`
= `[log|x|]_1^2 - [((x^(2 + 1))/(-2 + 1))]^2`
= `{log|2| - log|1|} - {1/x}_1^2`
= `[log 2 - 0] + [1/2 - 1/1]`
= `log 2 + [(1 - 2)/2]`
= `log 2 - 1/2`
= `1/2 [2 log 2 - 1]`
shaalaa.com
Definite Integrals
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]
\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]
\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]
\[\int\limits_1^2 \frac{x}{\left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx\] is equal to
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`Γ (9/2)`
