What is the integral from 2 to 3 of x^{2/3}(-1+x) ?

The integral from 2 to 3 of x^{2/3}(-1+x) is (63*3^{2/3})/(40)-3/(5*2^{1/3)}

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integral from 2 to 3 of x^{2/3}(-1+x)

\int_{2}^{3} x^{\frac{2}{3}} (- 1 + x) d x

\frac{63 \cdot 3 ^{\frac{2}{3}}}{40} - \frac{3}{5 \cdot 2 ^{\frac{1}{3}}}

Decimal

2.79991 \dots

Solution steps

\int_{2}^{3} x^{\frac{2}{3}} (- 1 + x) d x

Apply Integration By Parts

: [\frac{3}{5} x^{\frac{5}{3}} (- 1 + x) - \int \frac{3}{5} x^{\frac{5}{3}} d x]_{2}^{3}

= [\frac{3}{5} x^{\frac{5}{3}} (- 1 + x) - \int \frac{3}{5} x^{\frac{5}{3}} d x]_{2}^{3}

\int \frac{3}{5} x^{\frac{5}{3}} d x = \frac{9}{40} x^{\frac{8}{3}}

= [\frac{3}{5} x^{\frac{5}{3}} (- 1 + x) - \frac{9}{40} x^{\frac{8}{3}}]_{2}^{3}

Simplify

[\frac{3}{5} x^{\frac{5}{3}} (- 1 + x) - \frac{9}{40} x^{\frac{8}{3}}]_{2}^{3} : [\frac{1}{40} (- 24 x^{\frac{5}{3}} + 15 x^{\frac{8}{3}})]_{2}^{3}

= [\frac{1}{40} (- 24 x^{\frac{5}{3}} + 15 x^{\frac{8}{3}})]_{2}^{3}

Compute the boundaries:

\frac{63 \cdot 3 ^{\frac{2}{3}}}{40} - \frac{3}{5 \cdot 2 ^{\frac{1}{3}}}

= \frac{63 \cdot 3 ^{\frac{2}{3}}}{40} - \frac{3}{5 \cdot 2 ^{\frac{1}{3}}}

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\int_{- 2}^{3} (3 x^{4} - 5 x^{3} - x - 6) d x

\int_{0}^{1} x \cdot 1 + x^{2} d x

\int_{1}^{2} (- x^{2} + 5) d x

What is the integral from 2 to 3 of x^{2/3}(-1+x) ?
The integral from 2 to 3 of x^{2/3}(-1+x) is (63*3^{2/3})/(40)-3/(5*2^{1/3)}