Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_2^3 \frac{1}{x} d x\]
\[ = \left[ \log_e x \right]_2^3 \]
\[ = \log_e 3 - \log_e 2\]
\[ = \log_e \left( \frac{3}{2} \right)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If f(x) is a continuous function defined on [−a, a], then prove that
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
