What is the integral from 0 to 2 of sqrt(4-t^2) ?

The integral from 0 to 2 of sqrt(4-t^2) is pi

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integral from 0 to 2 of sqrt(4-t^2)

\int_{0}^{2} 4 - t^{2} d t

π

Decimal

3.14159 \dots

Solution steps

\int_{0}^{2} 4 - t^{2} d t

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 4 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 4 cos^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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

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

Simplify

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

= π

\int_{0}^{1} x^{2} + 3 x d x

\int_{1}^{2} 4 x^{5} d x

\int_{- 1}^{1} x x^{2} + 4 d x

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

\int_{0}^{\frac{π}{3}} sin (3 y) cos (y) d y

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