Advertisements
Advertisements
Question
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
Advertisements
Solution
We have,
\[\left| x^2 - 2x \right| = \begin{cases}- \left( x^2 - 2x \right),& 1 \leq x \leq 2\\ x^2 - 2x,& 2 \leq x \leq 3\end{cases}\]
\[ \therefore \int_1^3 \left| x^2 - 2x \right| d x\]
\[ = \int_1^2 - \left( x^2 - 2x \right) dx + \int_2^3 \left( x^2 - 2x \right) dx\]
\[ = \left[ - \frac{x^3}{3} + x^2 \right]_1^2 + \left[ \frac{x^3}{3} - x^2 \right]_2^3 \]
\[ = \frac{- 8}{3} + 4 + \frac{1}{3} - 1 + 9 - 9 - \frac{8}{3} + 4\]
= 2
APPEARS IN
RELATED QUESTIONS
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
Γ(1) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Find: `int logx/(1 + log x)^2 dx`
