009B Sample Midterm 2, Problem 2 Detailed Solution

Evaluate

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

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

Background Information:
How would you integrate  $\int (2x+1){\sqrt {x^{2}+x}}~dx?$
You can use  $u$-substitution.
Let  $u=x^{2}+x.$
Then,  $du=(2x+1)~dx.$
Thus,
${\begin{array}{rcl}\displaystyle {\int (2x+1){\sqrt {x^{2}+x}}~dx}&=&\displaystyle {\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{3}}u^{3/2}+C}\\&&\\&=&\displaystyle {{\frac {2}{3}}(x^{2}+x)^{3/2}+C.}\end{array}}$

Solution:

(a)

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

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

(b)

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

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

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