Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We know that
\[\left| x + 1 \right| = \begin{cases}x + 1, & \text{if }x + 1 \geq 0 \\ - \left( x + 1 \right), & \text{if }x + 1 < 0\end{cases} = \begin{cases}x + 1, & \text{if }x \geq - 1 \\ - \left( x + 1 \right), & \text{if }x < - 1\end{cases}\]
\[\left| x - 1 \right| = \begin{cases}x - 1, & \text{if }x - 1 \geq 0 \\ - \left( x - 1 \right), & \text{if }x - 1 < 0\end{cases} = \begin{cases}x - 1, & \text{if }x \geq 1 \\ - \left( x - 1 \right), & \text{if }x < 1\end{cases}\]
When
When
When
\[\therefore \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]
\[ = \int_{- 1}^0 \left( 2 - x \right)dx + \int_0^1 \left( x + 2 \right)dx + \int_1^2 3xdx\]
\[ = \left.\frac{\left( 2 - x \right)^2}{2 \times \left( - 1 \right)}\right|_{- 1}^0 + \left.\frac{\left( x + 2 \right)^2}{2}\right|_0^1 + \left.3 \times \frac{x^2}{2}\right|_1^2 \]
\[ = - \frac{1}{2}\left( 4 - 9 \right) + \frac{1}{2}\left( 9 - 4 \right) + \frac{3}{2}\left( 4 - 1 \right)\]
\[ = \frac{5}{2} + \frac{5}{2} + \frac{9}{2}\]
\[ = \frac{19}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
