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
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
