Question

Evaulate the definite integral (The s is the integration symbol with indicated values)

Answer 1

$\displaystyle\int\limits_{t=1}^{t=4}\frac{t^6-t^2}{t^4}\,\mathrm dt=\int_1^4\left(t^2-\frac1{t^2}\right)\,\mathrm dt=\left(\frac13t^3+\frac1t\right)\bigg|_{t=1}^{t=4}$

which evaluates to $\dfrac{259}{12}-\dfrac43=\dfrac{81}4$.