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

The integral from 3 to 7 of x\sqrt[3]{x+1} is (384-48*2^{2/3})/7-12+3\sqrt[3]{4}

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

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

\frac{384 - 48 \cdot 2 ^{\frac{2}{3}}}{7} - 12 + 334

Decimal

36.73431 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{8} (u - 1) 3 u d u

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

Expand

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

= \int_{4}^{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

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

\int_{4}^{8} u^{\frac{4}{3}} d u = \frac{384 - 48 \cdot 2 ^{\frac{2}{3}}}{7}

\int_{4}^{8} 3 u d u = 12 - 334

= \frac{384 - 48 \cdot 2 ^{\frac{2}{3}}}{7} - (12 - 334)

Simplify

= \frac{384 - 48 \cdot 2 ^{\frac{2}{3}}}{7} - 12 + 334

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What is the integral from 3 to 7 of x\sqrt[3]{x+1} ?
The integral from 3 to 7 of x\sqrt[3]{x+1} is (384-48*2^{2/3})/7-12+3\sqrt[3]{4}