Advertisements
Advertisements
प्रश्न
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
योग
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
Definite Integrals
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^2 \log\ x\ dx\]
\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]
\[\int\limits_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} dx\]
\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]
\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\] is equal to
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]