What is the integral from 0 to 4 of (13-t)sqrt(t) ?

The integral from 0 to 4 of (13-t)sqrt(t) is 848/15

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integral from 0 to 4 of (13-t)sqrt(t)

\int_{0}^{4} (13 - t) t d t

\frac{848}{15}

Solution steps

\int_{0}^{4} (13 - t) t d t

Expand

(13 - t) t : 13 t - t^{\frac{3}{2}}

= \int_{0}^{4} 13 t - t^{\frac{3}{2}} d t

Apply the Sum Rule:

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

= \int_{0}^{4} 13 t d t - \int_{0}^{4} t^{\frac{3}{2}} d t

\int_{0}^{4} 13 t d t = \frac{208}{3}

\int_{0}^{4} t^{\frac{3}{2}} d t = \frac{64}{5}

= \frac{208}{3} - \frac{64}{5}

Simplify

\frac{208}{3} - \frac{64}{5} : \frac{848}{15}

= \frac{848}{15}

y = t^{2} + 5 cos (t)

\frac{d y}{d x} sin (x + y) = 0

\int \frac{sec ( x ) + cos ( x )}{4 cos ( x )} d x

x \to 0 lim (x \cdot lo g_{2} (x))

\int \frac{x ^{2} + 36}{x} d x

What is the integral from 0 to 4 of (13-t)sqrt(t) ?
The integral from 0 to 4 of (13-t)sqrt(t) is 848/15