Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^2 2x\left[ x \right]dx\]
\[ = \int_0^1 2x\left[ x \right]dx + \int_1^2 2x\left[ x \right]dx\]
\[ = \int_0^1 2x \times 0dx + \int_1^2 2x \times 1dx .................\left[ \left[ x \right] = \begin{cases}0, & 0 \leq x < 1 \\ 1, & 1 \leq x < 2\end{cases} \right]\]
\[ = 0 + 2 \int_1^2 xdx\]
\[ = \left.2 \times \frac{x^2}{2}\right|_1^2 \]
\[ = 4 - 1\]
\[ = 3\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find: `int logx/(1 + log x)^2 dx`
