What is the integral from 0 to 1 of 8xsqrt(1-x^4) ?

The integral from 0 to 1 of 8xsqrt(1-x^4) is pi

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integral from 0 to 1 of 8xsqrt(1-x^4)

\int_{0}^{1} 8 x 1 - x^{4} d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{1} 8 x 1 - x^{4} d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x 1 - x^{4} d x

Simplify

x^{4} : (x^{2})^{2}

= 8 \cdot \int_{0}^{1} x 1 - (x^{2})^{2} d x

Apply u-substitution

: \int_{0}^{1} \frac{1 - u ^{2}}{2} d u

= 8 \cdot \int_{0}^{1} \frac{1 - u ^{2}}{2} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{1} 1 - u^{2} d u

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

Rewrite using trig identities

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 v) d v

Apply the Sum Rule:

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

= 8 \cdot \frac{1}{2} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v + \int_{0}^{\frac{π}{2}} cos (2 v) d v)

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= 8 \cdot \frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

8 \cdot \frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} + 0) : π

= π

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What is the integral from 0 to 1 of 8xsqrt(1-x^4) ?
The integral from 0 to 1 of 8xsqrt(1-x^4) is pi