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:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If f(2a − x) = −f(x), prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
