What is the integral from-1 to 1 of 1/4 x(2x+2) ?

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

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integral from-1 to 1 of 1/4 x(2x+2)

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{4} \cdot \int_{- 1}^{1} x (2 x + 2) d x

Apply Integration By Parts

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

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

\int x^{2} d x = \frac{x ^{3}}{3}

= \frac{1}{4} [x^{2} (x + 1) - \frac{x ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{4}{3}

= \frac{1}{4} \cdot \frac{4}{3}

Simplify

= \frac{1}{3}

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\int_{- 1}^{2} \frac{1}{3} x^{2} d x

\int_{0}^{l n (x)} (x e^{t^{2}}) d t

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