What is the integral from 0 to 7/2 of sqrt(49-x^2) ?

The integral from 0 to 7/2 of sqrt(49-x^2) is (49(2pi+3sqrt(3)))/(24)

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integral from 0 to 7/2 of sqrt(49-x^2)

\int_{0}^{\frac{7}{2}} 49 - x^{2} d x

\frac{49 ( 2 π + 3 3 )}{24}

Decimal Notation

23.43698 \dots

Solution steps

\int_{0}^{\frac{7}{2}} 49 - x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 49 cos^{2} (u) d u

= \int_{0}^{\frac{π}{6}} 49 cos^{2} (u) d u

Take the constant out:

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

= 49 \cdot \int_{0}^{\frac{π}{6}} cos^{2} (u) d u

Rewrite using trig identities

= 49 \cdot \int_{0}^{\frac{π}{6}} \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

= 49 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 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

= 49 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 49 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

Simplify

49 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : \frac{49 ( 2 π + 3 3 )}{24}

= \frac{49 ( 2 π + 3 3 )}{24}

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What is the integral from 0 to 7/2 of sqrt(49-x^2) ?
The integral from 0 to 7/2 of sqrt(49-x^2) is (49(2pi+3sqrt(3)))/(24)