What is the integral from-1 to 3 of (t+2)(t^2-2t+4) ?

The integral from-1 to 3 of (t+2)(t^2-2t+4) is 52

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integral from-1 to 3 of (t+2)(t^2-2t+4)

\int_{- 1}^{3} (t + 2) (t^{2} - 2 t + 4) d t

52

Solution steps

\int_{- 1}^{3} (t + 2) (t^{2} - 2 t + 4) d t

Apply Integration By Parts

= [(t + 2) (\frac{t ^{3}}{3} - t^{2} + 4 t) - \int \frac{t ^{3}}{3} - t^{2} + 4 t d t]_{- 1}^{3}

\int \frac{t ^{3}}{3} - t^{2} + 4 t d t = \frac{t ^{4}}{12} - \frac{t ^{3}}{3} + 2 t^{2}

= [(t + 2) (\frac{t ^{3}}{3} - t^{2} + 4 t) - (\frac{t ^{4}}{12} - \frac{t ^{3}}{3} + 2 t^{2})]_{- 1}^{3}

Simplify

[(t + 2) (\frac{t ^{3}}{3} - t^{2} + 4 t) - (\frac{t ^{4}}{12} - \frac{t ^{3}}{3} + 2 t^{2})]_{- 1}^{3} : [\frac{1}{4} t (t^{3} + 32)]_{- 1}^{3}

= [\frac{1}{4} t (t^{3} + 32)]_{- 1}^{3}

Compute the boundaries:

52

= 52

What is the integral from-1 to 3 of (t+2)(t^2-2t+4) ?
The integral from-1 to 3 of (t+2)(t^2-2t+4) is 52