How can I calculate $\int_{- 1}^{0} (x^{2} \cdot \sqrt{1 + x^{3}}) d x$ $int_-1^0(x^2*sqrt(1+x^3))dx$ ?

Question

How can I calculate $\int_{- 1}^{0} (x^{2} \cdot \sqrt{1 + x^{3}}) d x$ $int_-1^0(x^2*sqrt(1+x^3))dx$ ?

How can I calculate $\int_{- 1}^{0} (x^{2} \cdot \sqrt{1 + x^{3}}) d x$ $int_-1^0(x^2*sqrt(1+x^3))dx$

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Answer 1 · 2018-05-13T23:06:37

$\int_{- 1}^{0} x^{2} \sqrt{1 + x^{3}} = \frac{2}{9}$ $int_(-1)^0 x^2sqrt(1+x^3) = 2/9$

Explanation:

Note that:

$\frac{d}{d x} (1 + x^{3}) = 3 x^{2}$ $d/dx (1+x^3) = 3x^2$

so:

$\int_{- 1}^{0} x^{2} \sqrt{1 + x^{3}} = \frac{1}{3} \int_{- 1}^{0} \sqrt{1 + x^{3}} d (1 + x^{3})$ $int_(-1)^0 x^2sqrt(1+x^3) = 1/3 int_(-1)^0 sqrt(1+x^3) d(1+x^3)$

$\int_{- 1}^{0} x^{2} \sqrt{1 + x^{3}} = \frac{1}{3} {⎡ ⎢ ⎣ \frac{{(1 + x^{3})}^{\frac{3}{2}}}{\frac{3}{2}} ⎤ ⎥ ⎦}_{- 1}^{0}$ $int_(-1)^0 x^2sqrt(1+x^3) = 1/3 [(1+x^3)^(3/2)/(3/2)]_(-1)^0$

$\int_{- 1}^{0} x^{2} \sqrt{1 + x^{3}} = \frac{2}{9} [{(1 + 0)}^{\frac{3}{2}} - {(1 + {(- 1)}^{3})}^{\frac{3}{2}}] = \frac{2}{9}$ $int_(-1)^0 x^2sqrt(1+x^3) = 2/9 [(1+0)^(3/2)-(1+(-1)^3)^(3/2)] =2/9$

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How can I calculate $\int_{- 1}^{0} (x^{2} \cdot \sqrt{1 + x^{3}}) d x$ $int_-1^0(x^2*sqrt(1+x^3))dx$ ?