Question

use the fact that \\(\int_{a}^{b} x^2 \\, dx = \frac{b^3}{3} - \frac{a^3}{3}\\), where \\(a < b\\), to evaluate the integral \\(\int_{0}^{4.2} s^2 \\, ds\\).

the value of the integral \\(\int_{0}^{4.2} s^2 \\, ds = \square\\).

(type an exact answer in simplified form.)

Explanation:

Step1: Identify a and b

In the integral $\int_{0}^{4.2} s^2 ds$, compare with $\int_{a}^{b} x^2 dx$. So $a = 0$, $b = 4.2$.

Step2: Apply the formula

Use the formula $\int_{a}^{b} x^2 dx=\frac{b^3}{3}-\frac{a^3}{3}$. Substitute $a = 0$, $b = 4.2$ (which is $\frac{21}{5}$).
$\int_{0}^{4.2} s^2 ds=\frac{(4.2)^3}{3}-\frac{0^3}{3}=\frac{(4.2)^3}{3}$.
Calculate $(4.2)^3 = 4.2\times4.2\times4.2 = 74.088$. Then $\frac{74.088}{3}=24.696$. Or in fraction: $4.2=\frac{21}{5}$, so $(\frac{21}{5})^3=\frac{9261}{125}$, then $\frac{9261}{125\times3}=\frac{3087}{125}=24.696$.

Answer:

24.696 (or $\frac{3087}{125}$)