Advertisements
Advertisements
Question
Advertisements
Solution
\[\text{We have}, \]
\[I = \int\limits_0^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left[ x^2 \right] dx + \int\limits_1^\sqrt{2} \left[ x^2 \right] dx\]
\[ = \int\limits_0^1 \left( 0 \right)dx + \int\limits_1^\sqrt{2} \left( 1 \right)dx .................\left( \because \left[ x^2 \right] = \begin{cases}0&& 0 < x < 1\\1&& 1 < x < \sqrt{2}\end{cases} \right)\]
\[ = 0 + \left[ x \right]_1^\sqrt{2} \]
\[ = \sqrt{2} - 1\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Γ(1) is
