Answer
$\frac{17}{6}$
Work Step by Step
Evaluate the Integral: $\int^{2}_{1}\frac{s^4+1}{s^2}ds $
Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$
Find $F(x)$:
$F(x) = \int\frac{s^4+1}{s^2}ds$
$F(x) = \int\frac{s^4}{s^2}ds + \int\frac{1}{s^2}ds$
$F(x) = \int s^2ds + \int s^{-2}ds$
$F(x) = \frac{s^3}{3} - \frac{1}{s}$
$F(x) = \frac{s^4 - 3}{3s} $
Now Evaluate: $F(b) - F(a)$
$F(2) - F(1)$
$\frac{(2)^4 - 3}{3(2)} - \frac{(1)^4 -3}{3(1)}$
$\frac{16-3}{6} - \frac{1 - 3}{3}$
$\frac{13}{6} + \frac{2}{3}$
$\frac{17}{6}$
