Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
बेरीज
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
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]
\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]
\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]
\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]
Prove that:
\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
\[\int\limits_0^2 \left( x^2 + x \right) dx\]
\[\int\limits_1^2 \log_e \left[ x \right] dx .\]
\[\int\limits_0^4 x dx\]
`int x^3/(x + 1)` is equal to ______.
