What is the integral from 0 to 3 of |t^2+4t-5| ?

The integral from 0 to 3 of |t^2+4t-5| is 52/3

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integral from 0 to 3 of |t^2+4t-5|

\int_{0}^{3} t^{2} + 4 t - 5 d t

\frac{52}{3}

Decimal

17.33333 \dots

Solution steps

\int_{0}^{3} t^{2} + 4 t - 5 d t

Eliminate Absolutes

= \int_{0}^{1} - t^{2} - 4 t + 5 d t + \int_{1}^{3} t^{2} + 4 t - 5 d t

\int_{0}^{1} - t^{2} - 4 t + 5 d t = \frac{8}{3}

\int_{1}^{3} t^{2} + 4 t - 5 d t = \frac{44}{3}

= \frac{8}{3} + \frac{44}{3}

Simplify

\frac{8}{3} + \frac{44}{3} : \frac{52}{3}

= \frac{52}{3}

\int_{0}^{1} (x^{2} - x^{3})^{2} d x

\int_{- 2}^{0} x^{2} - 10 x d x

\int_{- π}^{π} t^{2} \cdot cos (n t) d t

\int_{0}^{l n (2)} 2 x^{2} e^{3 x} d x

\int_{- 1}^{2} e^{5 x - 4} d x

What is the integral from 0 to 3 of |t^2+4t-5| ?
The integral from 0 to 3 of |t^2+4t-5| is 52/3