What is the integral from-1 to 0 of (2t+2)^2 ?

The integral from-1 to 0 of (2t+2)^2 is 4/3

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integral from-1 to 0 of (2t+2)^2

\int_{- 1}^{0} (2 t + 2)^{2} d t

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

\int_{- 1}^{0} (2 t + 2)^{2} d t

Apply u-substitution

: \int_{0}^{2} \frac{u ^{2}}{2} d u

= \int_{0}^{2} \frac{u ^{2}}{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_{0}^{2} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{2}

= \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{2}

Compute the boundaries:

\frac{8}{3}

= \frac{1}{2} \cdot \frac{8}{3}

Simplify

= \frac{4}{3}

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What is the integral from-1 to 0 of (2t+2)^2 ?
The integral from-1 to 0 of (2t+2)^2 is 4/3