What is the integral from 0 to sqrt(5 of)sqrt(5-x^2) ?

The integral from 0 to sqrt(5 of)sqrt(5-x^2) is (5pi)/4

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integral from 0 to sqrt(5 of)sqrt(5-x^2)

\int_{0}^{5} 5 - x^{2} d x

\frac{5 π}{4}

Decimal Notation

3.92699 \dots

Solution steps

\int_{0}^{5} 5 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 5 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 5 cos^{2} (u) d u

Take the constant out:

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

= 5 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 5 \cdot \int_{0}^{\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

= 5 \cdot \frac{1}{2} \cdot \int_{0}^{\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

= 5 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 5 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

5 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{5 π}{4}

= \frac{5 π}{4}

\int_{0}^{2} x e^{\frac{- x ^{2} - 1}{2}} d x

\int_{1}^{4} - x - (2 - x) d x

\int_{0}^{h} \frac{b}{h} (h - y) d y

\int_{1}^{4} (2 x) d x

\int_{0}^{π} cos (\frac{1}{2}) x d x

What is the integral from 0 to sqrt(5 of)sqrt(5-x^2) ?
The integral from 0 to sqrt(5 of)sqrt(5-x^2) is (5pi)/4