Question

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evaluate the integral.
\\(\int_{0}^{3}\left(\frac{4}{3}t^{3} - \frac{3}{2}t^{2} + \frac{2}{3}t\
ight) dt\\)
\\(\frac{49}{2}\\) \\(\times\\)

Explanation:

Step1: Find antiderivative

Use power rule: $\int t^n dt = \frac{t^{n + 1}}{n + 1} + C$.
Antiderivative of $\frac{4}{3}t^3$: $\frac{4}{3} \cdot \frac{t^4}{4} = \frac{t^4}{3}$
Antiderivative of $-\frac{3}{2}t^2$: $-\frac{3}{2} \cdot \frac{t^3}{3} = -\frac{t^3}{2}$
Antiderivative of $\frac{2}{3}t$: $\frac{2}{3} \cdot \frac{t^2}{2} = \frac{t^2}{3}$
So antiderivative $F(t) = \frac{t^4}{3} - \frac{t^3}{2} + \frac{t^2}{3}$

Step2: Apply Fundamental Theorem

Evaluate $F(3) - F(0)$.
First, $F(3)$:
$\frac{3^4}{3} - \frac{3^3}{2} + \frac{3^2}{3} = \frac{81}{3} - \frac{27}{2} + \frac{9}{3} = 27 - \frac{27}{2} + 3$
Simplify: $30 - \frac{27}{2} = \frac{60}{2} - \frac{27}{2} = \frac{33}{2}$

$F(0) = \frac{0^4}{3} - \frac{0^3}{2} + \frac{0^2}{3} = 0$

Thus, integral is $\frac{33}{2} - 0 = \frac{33}{2}$

Answer:

$\frac{33}{2}$