What is the integral from 65 to 75 of (0.2x+8) ?

The integral from 65 to 75 of (0.2x+8) is 220

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integral from 65 to 75 of (0.2x+8)

\int_{65}^{75} (0.2 x + 8) d x

220

Solution steps

\int_{65}^{75} 0.2 x + 8 d x

Apply u-substitution

: \int_{21}^{23} \frac{u}{0.2} d u

= \int_{21}^{23} \frac{u}{0.2} d u

Take the constant out:

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

= \frac{1}{0.2} \cdot \int_{21}^{23} u d u

Simplify

= 5 \cdot \int_{21}^{23} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{21}^{23}

= 5 [\frac{u ^{2}}{2}]_{21}^{23}

Compute the boundaries:

44

= 5 \cdot 44

Simplify

= 220

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What is the integral from 65 to 75 of (0.2x+8) ?
The integral from 65 to 75 of (0.2x+8) is 220