What is the integral from 0 to 3 of sqrt(4-t) ?

The integral from 0 to 3 of sqrt(4-t) is 14/3

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integral from 0 to 3 of sqrt(4-t)

\int_{0}^{3} 4 - t d t

\frac{14}{3}

Decimal

4.66666 \dots

Solution steps

\int_{0}^{3} 4 - t d t

Apply u-substitution

: \int_{4}^{1} - u d u

= \int_{4}^{1} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{4} - u d u

Take the constant out:

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

= - (- \int_{1}^{4} u d u)

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

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

Simplify

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{4}

Compute the boundaries:

\frac{14}{3}

= \frac{14}{3}

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What is the integral from 0 to 3 of sqrt(4-t) ?
The integral from 0 to 3 of sqrt(4-t) is 14/3