What is the integral from 0 to 7 of 7x\sqrt[3]{x+1} ?

The integral from 0 to 7 of 7x\sqrt[3]{x+1} is 1209/4

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integral from 0 to 7 of 7x\sqrt[3]{x+1}

\int_{0}^{7} 7 x 3 x + 1 d x

\frac{1209}{4}

Decimal

302.25

Solution steps

\int_{0}^{7} 7 x 3 x + 1 d x

Take the constant out:

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

= 7 \cdot \int_{0}^{7} x 3 x + 1 d x

Apply u-substitution

: \int_{1}^{8} (u - 1) 3 u d u

= 7 \cdot \int_{1}^{8} (u - 1) 3 u d u

Expand

(u - 1) 3 u : u^{\frac{4}{3}} - 3 u

= 7 \cdot \int_{1}^{8} u^{\frac{4}{3}} - 3 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 7 (\int_{1}^{8} u^{\frac{4}{3}} d u - \int_{1}^{8} 3 u d u)

\int_{1}^{8} u^{\frac{4}{3}} d u = \frac{381}{7}

\int_{1}^{8} 3 u d u = \frac{45}{4}

= 7 (\frac{381}{7} - \frac{45}{4})

Simplify

7 (\frac{381}{7} - \frac{45}{4}) : \frac{1209}{4}

= \frac{1209}{4}

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What is the integral from 0 to 7 of 7x\sqrt[3]{x+1} ?
The integral from 0 to 7 of 7x\sqrt[3]{x+1} is 1209/4