What is the integral from-B to B of-sqrt(B^2-x^2) ?

The integral from-B to B of-sqrt(B^2-x^2) is -pi/2 B^2

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integral from-B to B of-sqrt(B^2-x^2)

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

- \frac{π}{2} B^{2}

Solution steps

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

Take the constant out:

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

= - \int_{- B}^{B} B^{2} - x^{2} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= - B^{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

= - B^{2} \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

= - B^{2} \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

= - B^{2} \frac{1}{2} (π + 0)

Simplify

- B^{2} \frac{1}{2} (π + 0) : - \frac{π}{2} B^{2}

= - \frac{π}{2} B^{2}

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What is the integral from-B to B of-sqrt(B^2-x^2) ?
The integral from-B to B of-sqrt(B^2-x^2) is -pi/2 B^2