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

解

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

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

解答ステップ

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

定数を除く:

\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

べき乗則を適用する

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

限度を計算する:

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

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

簡素化

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