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 the following integral:
Evaluate each of the following integral:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following:
`Γ (9/2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
`int x^3/(x + 1)` is equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?
