What is the integral from-1 to 1 of 2sqrt(1-y^2) ?

The integral from-1 to 1 of 2sqrt(1-y^2) is pi

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integral from-1 to 1 of 2sqrt(1-y^2)

\int_{- 1}^{1} 2 1 - y^{2} d y

π

Decimal

3.14159 \dots

Solution steps

\int_{- 1}^{1} 2 1 - y^{2} d y

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} 1 - y^{2} d y

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

= 2 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 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

= 2 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u = π

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

= 2 \cdot \frac{1}{2} (π + 0)

Simplify

2 \cdot \frac{1}{2} (π + 0) : π

= π

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What is the integral from-1 to 1 of 2sqrt(1-y^2) ?
The integral from-1 to 1 of 2sqrt(1-y^2) is pi