Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
योग
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]
\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]
\[\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx\]
\[\int_0^1 | x\sin \pi x | dx\]
\[\int\limits_1^3 \left( 3x - 2 \right) dx\]
Evaluate each of the following integral:
\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
