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

The integral from-7 to 0 of sqrt(49-x^2) is (49pi)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{49 π}{4}

Decimal Notation

38.48451 \dots

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

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

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

= 49 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplify

49 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{49 π}{4}

= \frac{49 π}{4}

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

\int_{0}^{\frac{π}{2}} cos (x) d x

\int_{1}^{3} \frac{x ^{3} - 1}{x ^{4}} d x

\int_{0}^{l n (3)} 9 e^{3 x} d x

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

What is the integral from-7 to 0 of sqrt(49-x^2) ?
The integral from-7 to 0 of sqrt(49-x^2) is (49pi)/4