Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_1^2 \log\ x\ d\ x\ . Then, \]
\[I = \int_1^2 1 \log x\ d\ x\]
\[\text{Integrating by parts}\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 \frac{1}{x} x\ d\ x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \int_1^2 d x\]
\[ \Rightarrow I = \left[ x \log x \right]_1^2 - \left[ x \right]_1^2 \]
\[ \Rightarrow I = 2 \log 2 - 2 + 1\]
\[ \Rightarrow I = 2 \log 2 - 1\]
APPEARS IN
RELATED QUESTIONS
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
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`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
