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

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

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Simplify

(2 t)^{2} : 4 t^{2}

= \int_{0}^{1} 4 t^{2} d t

Take the constant out:

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

= 4 \cdot \int_{0}^{1} t^{2} d t

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{0}^{1}

= 4 [\frac{t ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

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

Simplify

= \frac{4}{3}

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