What is the integral from-1 to 4 of x(x^2-3)^5 ?

The integral from-1 to 4 of x(x^2-3)^5 is 1608915/4

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integral from-1 to 4 of x(x^2-3)^5

\int_{- 1}^{4} x (x^{2} - 3)^{5} d x

\frac{1608915}{4}

Decimal

402228.75

Solution steps

\int_{- 1}^{4} x (x^{2} - 3)^{5} d x

Apply u-substitution

: \int_{- 2}^{13} \frac{u ^{5}}{2} d u

= \int_{- 2}^{13} \frac{u ^{5}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- 2}^{13} u^{5} d u

Apply the Power Rule

: [\frac{u ^{6}}{6}]_{- 2}^{13}

= \frac{1}{2} [\frac{u ^{6}}{6}]_{- 2}^{13}

Compute the boundaries:

\frac{1608915}{2}

= \frac{1}{2} \cdot \frac{1608915}{2}

Simplify

= \frac{1608915}{4}

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What is the integral from-1 to 4 of x(x^2-3)^5 ?
The integral from-1 to 4 of x(x^2-3)^5 is 1608915/4