What is the integral from 0 to sin(x) of 8sqrt(t) ?

The integral from 0 to sin(x) of 8sqrt(t) is 16/3 sin^{3/2}(x)

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integral from 0 to sin(x) of 8sqrt(t)

\int_{0}^{s i n (x)} 8 t d t

\frac{16}{3} sin^{\frac{3}{2}} (x)

Solution steps

\int_{0}^{s i n (x)} 8 t d t

Take the constant out:

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

= 8 \cdot \int_{0}^{s i n (x)} t d t

Apply the Power Rule

: [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{s i n (x)}

= 8 [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{s i n (x)}

Compute the boundaries:

\frac{2}{3} sin^{\frac{3}{2}} (x)

= 8 \cdot \frac{2}{3} sin^{\frac{3}{2}} (x)

Simplify

= \frac{16}{3} sin^{\frac{3}{2}} (x)

\int - x - 3 d x

h (x) = (x^{2} - 3)^{3}

\int_{- 8}^{2} \frac{1}{∣ 2 x ∣} d x

\frac{d}{d y} (sin (2 y))

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 0

What is the integral from 0 to sin(x) of 8sqrt(t) ?
The integral from 0 to sin(x) of 8sqrt(t) is 16/3 sin^{3/2}(x)