d express your answer in simplest form. \n\n int \frac{15x^{2}sqrt4{4 - x^{3}}}{4} dx

Explanation:

Step1: Use substitution

Let $u = 4 - x^{3}$, then $\mathrm{d}u=-3x^{2}\mathrm{d}x$, and $x^{2}\mathrm{d}x=-\frac{1}{3}\mathrm{d}u$.

Step2: Rewrite the integral

The integral $\int\frac{15x^{2}\sqrt[4]{4 - x^{3}}}{4}\mathrm{d}x$ becomes $\int\frac{15}{4}\sqrt[4]{u}\left(-\frac{1}{3}\right)\mathrm{d}u=-\frac{5}{4}\int u^{\frac{1}{4}}\mathrm{d}u$.

Step3: Integrate using power - rule

The power - rule for integration is $\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). So, $-\frac{5}{4}\int u^{\frac{1}{4}}\mathrm{d}u=-\frac{5}{4}\times\frac{u^{\frac{1}{4}+1}}{\frac{1}{4}+1}+C$.

Step4: Simplify the result

$-\frac{5}{4}\times\frac{u^{\frac{5}{4}}}{\frac{5}{4}}+C=-u^{\frac{5}{4}}+C$.

Step5: Substitute back $u = 4 - x^{3}$

The final result is $-(4 - x^{3})^{\frac{5}{4}}+C$.

Answer:

$-(4 - x^{3})^{\frac{5}{4}}+C$