Advertisements
Advertisements
Question
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
Advertisements
Solution
\[\int_1^3 \left| x^2 - 4 \right| d x\]
\[ = \int_1^2 - \left( x^2 - 4 \right) dx + \int_2^3 \left( x^2 - 4 \right) dx\]
\[ = \left[ - \frac{x^3}{3} + 4x \right]_1^2 + \left[ \frac{x^3}{3} - 4x \right]_2^3 \]
\[ = \frac{- 8}{3} + 8 + \frac{1}{3} - 4 + 9 - 12 - \frac{8}{3} + 8\]
\[ = 4\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
