Advertisements
Advertisements
Question
Advertisements
Solution
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
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If f is an integrable function, show that
Evaluate each of the following integral:
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
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(4)
