Question

find the exact value.
\\(\int_{0}^{5} \left(x^{\frac{3}{2}}\
ight) dx = ?\sqrt{\quad}\\)

Explanation:

Step1: Apply Power Rule for Integration

The power rule for integration states that $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ (where $n
eq - 1$). For the function $f(x)=x^{\frac{3}{2}}$, we have $n = \frac{3}{2}$. So, the antiderivative $F(x)$ is $\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}=\frac{x^{\frac{5}{2}}}{\frac{5}{2}}=\frac{2}{5}x^{\frac{5}{2}}$.

Step2: Evaluate Definite Integral

Using the fundamental theorem of calculus, $\int_{a}^{b}f(x)dx=F(b)-F(a)$. Here, $a = 0$ and $b = 5$. So, $F(5)-F(0)=\frac{2}{5}(5)^{\frac{5}{2}}-\frac{2}{5}(0)^{\frac{5}{2}}$. Since $0^{\frac{5}{2}} = 0$, we have $\frac{2}{5}(5)^{\frac{5}{2}}$. We can rewrite $5^{\frac{5}{2}}$ as $5^{2+\frac{1}{2}}=5^{2}\times5^{\frac{1}{2}} = 25\sqrt{5}$. Then $\frac{2}{5}\times25\sqrt{5}=10\sqrt{5}$.

Answer:

$10\sqrt{5}$ (So the first box is $10$ and the second box is $5$)