Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
योग
Advertisements
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \frac{x}{x + 1} dx\]
\[\int\limits_0^\pi x \cos^2 x\ dx\]
Evaluate the following integral:
\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]
Evaluate the following integral:
\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]
\[\int\limits_1^4 \left( x^2 - x \right) dx\]
\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
