Advertisements
Advertisements
Question
Find : `∫_a^b logx/x` dx
Advertisements
Solution
Put `log x = t ⇒ 1/x dx = dt`
⇒ `x = a ⇒ t = loga & x = b ⇒ t = log b`
`therefore I = ∫_log a ^log b t dt`
= `t^2/2|_log a^log b`
= `1/2 [(log b)^2 - (log a)^2]`
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
Solve each of the following integral:
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
