What is the integral from 0 to 2.5 of 80x ?

The integral from 0 to 2.5 of 80x is 250

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integral from 0 to 2.5 of 80x

\int_{0}^{2.5} 80 x d x

250

Solution steps

\int_{0}^{2.5} 80 x d x

Take the constant out:

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

= 80 \cdot \int_{0}^{2.5} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{2.5}

= 80 [\frac{x ^{2}}{2}]_{0}^{2.5}

Simplify

= 80 [0.5 x^{2}]_{0}^{2.5}

Compute the boundaries:

3.125

= 80 \cdot 3.125

Simplify

= 250

\int 9 sin^{3} (x co) s^{5} x d x

\int - \frac{1}{sec ^{2} ( x )} d x

\int \frac{x}{( x ^{2} + 3 ) ^{2}} d x

f (x) = 12 x - 1.86 x^{2}, at x = a

y = (3 x + 1) 8 x - 1

What is the integral from 0 to 2.5 of 80x ?
The integral from 0 to 2.5 of 80x is 250