What is the integral from 0 to 1 of 9t ?

The integral from 0 to 1 of 9t is 9/2

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integral from 0 to 1 of 9t

\int_{0}^{1} 9 t d t

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{0}^{1} 9 t d t

Take the constant out:

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

= 9 \cdot \int_{0}^{1} t d t

Apply the Power Rule

: [\frac{t ^{2}}{2}]_{0}^{1}

= 9 [\frac{t ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= 9 \cdot \frac{1}{2}

Simplify

= \frac{9}{2}

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