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
\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]
\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]
\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]
\[\int\limits_0^\infty \frac{\log x}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
