What is the integral from-1 to 1 of pi(7-7x^2)^2 ?

The integral from-1 to 1 of pi(7-7x^2)^2 is pi 784/15

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integral from-1 to 1 of pi(7-7x^2)^2

\int_{- 1}^{1} π (7 - 7 x^{2})^{2} d x

π \frac{784}{15}

Decimal

164.20057 \dots

Solution steps

\int_{- 1}^{1} π (7 - 7 x^{2})^{2} d x

Take the constant out:

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

= π \cdot \int_{- 1}^{1} (7 - 7 x^{2})^{2} d x

Expand

(7 - 7 x^{2})^{2} : 49 - 98 x^{2} + 49 x^{4}

= π \cdot \int_{- 1}^{1} 49 - 98 x^{2} + 49 x^{4} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= π (\int_{- 1}^{1} 49 d x - \int_{- 1}^{1} 98 x^{2} d x + \int_{- 1}^{1} 49 x^{4} d x)

\int_{- 1}^{1} 49 d x = 98

\int_{- 1}^{1} 98 x^{2} d x = \frac{196}{3}

\int_{- 1}^{1} 49 x^{4} d x = \frac{98}{5}

= π (98 - \frac{196}{3} + \frac{98}{5})

Join

98 - \frac{196}{3} + \frac{98}{5} : \frac{784}{15}

= π \frac{784}{15}

\frac{d}{d x} (\frac{8 x ^{2} + 6 x + 2}{x})

\frac{\partial}{\partial x} (55 x - 0.5 x^{2})

\int \frac{- x - 1}{x ^{2} + 9} d x

\frac{t d y}{d t} + 3 y = t^{4}, t > 0, y (1) = 3

f (x) = x^{3} - 3 x, (2, 2)

What is the integral from-1 to 1 of pi(7-7x^2)^2 ?
The integral from-1 to 1 of pi(7-7x^2)^2 is pi 784/15