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

The integral from-1 to 1 of (2-x^2)^{3/2} is (12pi+32)/(16)

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

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

\frac{12 π + 32}{16}

Decimal Notation

4.35619 \dots

Solution steps

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

Apply Integral Substitution

= \int_{- 2}^{2} \frac{( 8 - u ^{2} ) ^{\frac{3}{2}}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{- 2}^{2} (8 - u^{2})^{\frac{3}{2}} d u

Apply Integration By Parts

: [u (8 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u]_{- 2}^{2}

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - \int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u]_{- 2}^{2}

\int - 3 u^{2} (8 - u^{2})^{\frac{1}{2}} d u = - 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} - (- 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}))))]_{- 2}^{2}

Simplify

= \frac{1}{16} [u (8 - u^{2})^{\frac{3}{2}} + 3 (\frac{1}{4} u^{3} (- u^{2} + 8)^{\frac{1}{2}} + 8 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}}) - 4 sin (2 arcsin (\frac{u}{2 \cdot 2 ^{\frac{1}{2}}})))]_{- 2}^{2}

Compute the boundaries:

12 π + 32

= \frac{1}{16} (12 π + 32)

Simplify

= \frac{12 π + 32}{16}

\int_{0}^{r} (a r^{3} + b r) 2 π r d r

\int_{- 1}^{1} \frac{1}{2} x d x

\int_{2}^{\infty} x^{- 2} d x

\int_{0}^{1} \frac{( - x + 1 ) ^{2}}{2} d x

\int_{0}^{1} c y^{2} (1 - y)^{4} d y

What is the integral from-1 to 1 of (2-x^2)^{3/2} ?
The integral from-1 to 1 of (2-x^2)^{3/2} is (12pi+32)/(16)