Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
बेरीज
Advertisements
उत्तर
`int_1^2 (x "d"x)/(x^2 + 1) = 1/2 int_1^2 (2xdx)/(x^2 + 1)`
= `1/2 int_1^2 ("d"(x^2 + 1))/(x^2 + 1)`
=`1/2 [log|x^2 + 1|]_1^2`
= `1/2 [log|2^2 + 1| - log|1^2+ 1|]`
= `1/2 [log 5 - log 2]`
= `1/2 log[5/2]` .......`{"Using" log "a" - log "b" = log ("a"/"b")}`
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]
\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]
\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]
\[\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx\]
\[\int\limits_0^\pi x \log \sin x\ dx\]
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals
\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
