What is the integral from 1 to 2 of x^3(x^2-4) ?

The integral from 1 to 2 of x^3(x^2-4) is -9/2

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integral from 1 to 2 of x^3(x^2-4)

\int_{1}^{2} x^{3} (x^{2} - 4) d x

- \frac{9}{2}

Decimal Notation

- 4.5

Solution steps

\int_{1}^{2} x^{3} (x^{2} - 4) d x

Apply Integration By Parts

: [\frac{1}{4} x^{4} (x^{2} - 4) - \int \frac{x ^{5}}{2} d x]_{1}^{2}

= [\frac{1}{4} x^{4} (x^{2} - 4) - \int \frac{x ^{5}}{2} d x]_{1}^{2}

\int \frac{x ^{5}}{2} d x = \frac{x ^{6}}{12}

= [\frac{1}{4} x^{4} (x^{2} - 4) - \frac{x ^{6}}{12}]_{1}^{2}

Simplify

[\frac{1}{4} x^{4} (x^{2} - 4) - \frac{x ^{6}}{12}]_{1}^{2} : [\frac{1}{6} x^{4} (x^{2} - 6)]_{1}^{2}

= [\frac{1}{6} x^{4} (x^{2} - 6)]_{1}^{2}

Compute the boundaries:

- \frac{9}{2}

= - \frac{9}{2}

\frac{10}{s ( s + 50 )}

y = 2 x, y = x

\frac{\partial}{\partial x} (sin (4 x^{2}))

\int - 3 x e^{x} d x

\int \frac{3}{x} + x d x

What is the integral from 1 to 2 of x^3(x^2-4) ?
The integral from 1 to 2 of x^3(x^2-4) is -9/2