# 007B Sample Midterm 2, Problem 2 Detailed Solution

Evaluate

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

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

Background Information:
How would you integrate  ${\displaystyle \int (2x+1){\sqrt {x^{2}+x}}~dx?}$

You can use  ${\displaystyle u}$-substitution.

Let  ${\displaystyle u=x^{2}+x.}$
Then,  ${\displaystyle du=(2x+1)~dx.}$

Thus,

${\displaystyle {\begin{array}{rcl}\displaystyle {\int (2x+1){\sqrt {x^{2}+x}}~dx}&=&\displaystyle {\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{3}}u^{3/2}+C}\\&&\\&=&\displaystyle {{\frac {2}{3}}(x^{2}+x)^{3/2}+C.}\end{array}}}$

Solution:

(a)

Step 1:
We multiply the product inside the integral to get

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt}&=&\displaystyle {\int _{1}^{2}{\bigg (}8t^{3}-10+12-{\frac {15}{t^{3}}}{\bigg )}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}(8t^{3}+2-15t^{-3})~dt.}\end{array}}}$

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt}&=&\displaystyle {2(2)^{4}+2(2)+{\frac {15}{2(2)^{2}}}-{\bigg (}2+2+{\frac {15}{2}}{\bigg )}}\\&&\\&=&\displaystyle {36+{\frac {15}{8}}-4-{\frac {15}{2}}}\\&&\\&=&\displaystyle {{\frac {211}{8}}.}\end{array}}}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx}&=&\displaystyle {{\frac {1}{4}}\int _{4}^{28}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {\left.{\frac {1}{6}}u^{\frac {3}{2}}\right|_{4}^{28}}\\&&\\&=&\displaystyle {{\frac {1}{6}}(28^{\frac {3}{2}}-4^{\frac {3}{2}})}\\&&\\&=&\displaystyle {{\frac {1}{6}}(({\sqrt {28}})^{3}-({\sqrt {4}})^{3})}\\&&\\&=&\displaystyle {{\frac {1}{6}}((2{\sqrt {7}})^{3}-2^{3}).}\end{array}}}$

Therefore,
${\displaystyle \int _{0}^{2}(x^{3}+x){\sqrt {x^{4}+2x^{2}+4}}~dx={\frac {28{\sqrt {7}}-4}{3}}.}$

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