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 the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
