Answer
$\frac{82}{5}$
Work Step by Step
Evaluate the integral: $\int^{4}_{1}\frac{2+x^2}{\sqrt x}dx$
Recall the 2nd part of the Fundamental Theorem of Calculus:
$∫^b_af(x)dx=F(b)−F(a)$
Find $F(x)$:
$F(x)=\int\frac{2+x^2}{\sqrt x}dx$
$F(x)=\int (\frac{2}{\sqrt x} + \frac{x^2}{\sqrt x})dx$
$F(x)=\int 2x^{-1/2} + x^2(x^{-1/2})dx$
$F(x)=\int 2x^{-1/2} + x^{3/2}dx$
$F(x)=4x^{1/2} + \frac{2}{5}x^{5/2}$
$F(x)=4\sqrt x + \frac{2}{5}{\sqrt x}^5$
Now evaluate $F(b)−F(a)$:
$F(4)−F(1)$
$(4\sqrt 4 + \frac{2}{5}{\sqrt 4}^5) - (4\sqrt 1 + \frac{2}{5}{\sqrt 1}^5)$
$(8+ \frac{2}{5}(32))-(4 + \frac{2}{5})$
$4 + \frac{62}{5}$
$\frac{82}{5}$
