int_0^a {{x^2}sin {x^3},dx} equals | vedclass

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(D) Let $I = \int_0^a {{x^2}\sin {x^3}\,dx}$.Substitute $t = x^3$,then $dt = 3x^2\,dx$,which implies $x^2\,dx = \frac{dt}{3}$.When $x = 0$,$t = 0^3 =

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$\frac{1}{3}(1 - \cos {a^3})$

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(D) Let $I = \int_0^a {{x^2}\sin {x^3}\,dx}$.Substitute $t = x^3$,then $dt = 3x^2\,dx$,which implies $x^2\,dx = \frac{dt}{3}$.When $x = 0$,$t = 0^3 = 0$. When $x = a$,$t = a^3$.Substituting these into the integral:$I = \int_0^{a^3} \sin(t) \cdot \frac{dt}{3} = \frac{1}{3} \int_0^{a^3} \sin(t)\,dt$.Integrating $\sin(t)$ gives $-\cos(t)$:$I = \frac{1}{3} [-\cos(t)]_0^{a^3} = -\frac{1}{3} [\cos(a^3) - \cos(0)]$.Since $\cos(0) = 1$,we have:$I = -\frac{1}{3} [\cos(a^3) - 1] = \frac{1}{3} (1 - \cos(a^3))$.

$\int_0^a {{x^2}\sin {x^3}\,dx} $ equals

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