What is the integral from 0 to 5 of 9800pi(5-y) ?

The integral from 0 to 5 of 9800pi(5-y) is 122500pi

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integral from 0 to 5 of 9800pi(5-y)

\int_{0}^{5} 9800 π (5 - y) d y

122500 π

Decimal Notation

384845.10006 \dots

Solution steps

\int_{0}^{5} 9800 π (5 - y) d y

Take the constant out:

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

= 9800 π \cdot \int_{0}^{5} 5 - y d y

Apply the Sum Rule:

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

= 9800 π (\int_{0}^{5} 5 d y - \int_{0}^{5} y d y)

\int_{0}^{5} 5 d y = 25

\int_{0}^{5} y d y = \frac{25}{2}

= 9800 π (25 - \frac{25}{2})

Simplify

9800 π (25 - \frac{25}{2}) : 122500 π

= 122500 π

\int (sin (x) cos^{3} (x)) d x

x \to \infty lim (\frac{( 2 x - 1 ) ( 3 - x )}{( x - 1 ) ( x + 3 )})

y^{^{'}} = \frac{2 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

\int - \frac{sin ( 2 x )}{4} d x

y^{^{''}} + y^{^{'}} + y = cos (t)

What is the integral from 0 to 5 of 9800pi(5-y) ?
The integral from 0 to 5 of 9800pi(5-y) is 122500pi