What is the integral from 0 to sin(x) of 6sqrt(t) ?

The integral from 0 to sin(x) of 6sqrt(t) is 4sin^{3/2}(x)

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integral from 0 to sin(x) of 6sqrt(t)

\int_{0}^{s i n (x)} 6 t d t

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

Solution steps

\int_{0}^{s i n (x)} 6 t d t

Take the constant out:

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

= 6 \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)}

= 6 [\frac{2}{3} t^{\frac{3}{2}}]_{0}^{s i n (x)}

Compute the boundaries:

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

= 6 \cdot \frac{2}{3} sin^{\frac{3}{2}} (x)

Simplify

= 4 sin^{\frac{3}{2}} (x)

y^{^{''}} + 2 y^{^{'}} - 48 y = 0

\int \frac{u}{u} d u

f (x) = \frac{1}{x} + 4, (- 2, 8)

\frac{\partial}{\partial x} (3 t^{2} e^{4 x})

x \to 1 lim ((3 x - 1)^{2})

What is the integral from 0 to sin(x) of 6sqrt(t) ?
The integral from 0 to sin(x) of 6sqrt(t) is 4sin^{3/2}(x)