007B 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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int (2x+1){\sqrt {x^{2}+x}}~dx?}
You can use  $u$-substitution.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x^{2}+x.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=(2x+1)~dx.}
Thus,
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
${\begin{array}{rcl}\displaystyle {\int _{1}^{2}{\bigg (}2t+{\frac {3}{t^{2}}}{\bigg )}{\bigg (}4t^{2}-{\frac {5}{t}}{\bigg )}~dt}&=&\displaystyle {2(2)^{4}+2(2)+{\frac {15}{2(2)^{2}}}-{\bigg (}2+2+{\frac {15}{2}}{\bigg )}}\\&&\\&=&\displaystyle {36+{\frac {15}{8}}-4-{\frac {15}{2}}}\\&&\\&=&\displaystyle {{\frac {211}{8}}.}\end{array}}$

(b)

Step 1:
We use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^4+2x^2+4.}
Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=(4x^3+4x)dx}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{4}=(x^3+x)dx.}
Also, we need to change the bounds of integration.
Plugging in our values into the equation  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x^4+2x^2+4,}   we get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=0^4+2(0)^2+4=4}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=2^4+2(2)^2+4=28.}
Therefore, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{4}\int_4^{28}\sqrt{u}~du.}

Step 2:
We now have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.}

Final Answer:
(a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{211}{8}}
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{28\sqrt{7}-4}{3}}

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