009B Sample Midterm 2, Problem 3

Evaluate

a) $\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt$

b) $\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx$

Foundations:
How would you integrate $\int (2x+1){\sqrt {x^{2}+x}}~dx?$
You could use $u$-substitution. Let $u=x^{2}+x.$ Then, $du=(2x+1)~dx.$
Thus, $\int (2x+1){\sqrt {x^{2}+x}}~dx=\int {\sqrt {u}}={\frac {2}{3}}u^{3/2}+C={\frac {2}{3}}(x^{2}+x)^{3/2}+C.$

Solution:

(a)

Step 1:
We multiply the product inside the integral to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\int _{1}^{2}{\bigg (}8t^{3}-10+12-{\frac {15}{t^{3}}}{\bigg )}~dt=\int _{1}^{2}(8t^{3}+2-15t^{-3})~dt$.

Step 2:
We integrate to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=\left.2t^{4}+2t+{\frac {15}{2}}t^{-2}\right|_{1}^{2}$.
We now evaluate to get
$\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt=2(2)^{4}+2(2)+{\frac {15}{2(2)^{2}}}-{\bigg (}2+2+{\frac {15}{2}}{\bigg )}=36+{\frac {15}{8}}-4-{\frac {15}{2}}={\frac {211}{8}}$.

(b)

Step 1:
We use $u$-substitution. Let $u=x^{4}+2x^{2}+4$. Then, $du=(4x^{3}+4x)dx$ and ${\frac {du}{4}}=(x^{3}+x)dx$. Also, we need to change the bounds of integration.
Plugging in our values into the equation $u=x^{4}+2x^{2}+4$, we get $u_{1}=0^{4}+2(0)^{2}+4=4$ and $u_{2}=2^{4}+2(2)^{2}+4=28$.
Therefore, the integral becomes  ${\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du$.

Step 2:
We now have:
$\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du=\left.{\frac {1}{6}}u^{\frac {3}{2}}\right|_{4}^{28}={\frac {1}{6}}(28^{\frac {3}{2}}-4^{\frac {3}{2}})={\frac {1}{6}}(({\sqrt {28}})^{3}-({\sqrt {4}})^{3})={\frac {1}{6}}((2{\sqrt {7}})^{3}-2^{3})$.
So, we have
$\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {28{\sqrt {7}}-4}{3}}$.

Final Answer:
(a)   ${\frac {211}{8}}$
(b)   ${\frac {28{\sqrt {7}}-4}{3}}$

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