Henrik Skov's copy of mathcentre: Definite integration

Question 1

Evaluate the following definite integrals.

a)

\[I=\int_1^{\var{b1}}\simplify[std]{({a1} * x ^ 2 + {c1} * x + {d1})^2}\;dx\]

$I=\;\;$Expected answer:

Input your answer to 2 decimal places.

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b)

\[I=\int_0^{\var{b2}}\simplify[std]{1/(x+{m2})}\;dx\]

$I=\;\;$Expected answer:

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c)

\[I=\int_0^\pi\simplify[std]{x * ({w} * Sin({m3} * x) + {1 -w} * Cos({m3} * x))}\;dx\]

$I=\;\;$Expected answer:

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d)

\[I=\int_0^{\var{b4}}\simplify[std]{x ^ {m4} * Exp({n4} * x)}\;dx\]

$I=\;\;$Expected answer:

Input your answer to 4 decimal places.

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a)
\[I=\int_1^\var{b1}\simplify[std]{({a1} * x ^ 2 + {c1} * x + {d1})^2}\;dx\]
Expand the parentheses to obtain:

\[\begin{eqnarray*}I &=& \int_1^\var{b1} \simplify[std]{{a1 ^ 2} * x ^ 4 + {2 * a1 * c1} * x ^ 3+ {c1 ^ 2+2*a1*d1} * x ^ 2 + {2 * c1 * d1} * x+ {d1 ^ 2} }\;dx\\ &=&\left[\simplify[std]{{a1 ^ 2}/5 * x ^ 5 + {2 * a1 * c1}/4 * x ^ 4+ {c1 ^ 2+2*a1*d1}/3 * x ^ 3 + {2 * c1 * d1}/2 * x^2+ {d1 ^ 2}x }\right]_1^\var{b1}\\ &=&\var{tans1}\\ \\&=&\var{ans1}\mbox{ to 2 decimal places} \end{eqnarray*} \]

b)
\[\begin{eqnarray*}I&=&\int_0^{\var{b2}}\simplify[std]{1/(x+{m2})}\;dx\\ &=&\left[\ln(x+\var{m2})\right]_0^{\var{b2}}\\ &=& \ln(\var{b2+m2})-\ln(\var{m2})\\ &=&\ln\left(\frac{\var{b2+m2}}{\var{m2}}\right)\\ &=&\var{ans2}\mbox{ to 2 decimal places} \end{eqnarray*} \]

c)
\[I=\int_0^\pi\simplify[std]{x * ({w} * Sin({m3} * x) + {1 -w} * Cos({m3} * x))}\;dx\]
We use integration by parts.

Recall that:
\[\int u\frac{dv}{dx}\;dx=uv-\int \frac{du}{dx}\;v\;dx\]
Here we set $u=x$ and $\displaystyle \frac{dv}{dx}=\simplify[std]{ {w} * Sin({m3} * x) + {1 -w} * Cos({m3} * x)}$

Hence \[v=\simplify[std]{({-w}/ {m3}) * Cos({m3} * x) + {1 -w} * (({1-w}/ {m3}) * Sin({m3} * x))}\]

So \[\begin{eqnarray*} I&=&\left[\simplify[std]{{-w}*((x / {m3}) * Cos({m3} * x)) + {1 -w} * ((x / {m3}) * Sin({m3} * x))}\right]_0^\pi -\int_0^\pi\simplify[std]{ ({ -w} / {m3} )* Cos({m3} * x) + {1 -w} * (1 / {m3} * Sin({m3} * x))}\;dx\\ &=&\simplify[std]{({-w*cos(m3*pi)})*({pi}/{m3})}-\left[\simplify[std]{{ -w} * (1 / {m3 ^ 2})* Sin({m3} * x) -({1 -w} * (1 / {m3 ^ 2}) * Cos({m3} * x))}\right]_0^\pi\\ &=& \var{ans3}\mbox{ to 2 decimal places} \end{eqnarray*} \]
d)

\[I=\int_0^{\var{b4}}\simplify[std]{x ^ {m4} * Exp({n4} * x)}\;dx\]

Use integration by parts twice with $u=x^2$ and $\displaystyle \frac{dv}{dx}=\simplify[std]{e^({n4}x)}\Rightarrow v = \simplify[std]{1/{n4}e^({n4}x)}$
\[\begin{eqnarray*} I&=&\left[\simplify[std]{x^2/{n4}Exp({n4} * x)}\right]_0^{\var{b4}}+\simplify[std]{2/{abs(n4)}DefInt(x*Exp({n4} * x),x,0,{b4})}\\ &=&\simplify[std]{{b4^2}/{n4}*e^{p}-2/{n4}}\left(\left[\simplify[std]{x/{n4}*e^({n4}*x)}\right]_0^{\var{b4}}+\simplify[std]{1/{abs(n4)}DefInt(e^({n4}x),x,0,{b4})}\right)\\ &=&\simplify[std]{{b4^2}/{n4}*e^{p}-2/{n4}}\left(\simplify[std]{{b4}/{n4}*e^{p}-1/{n4}}\left[\simplify[std]{1/{n4}*e^({n4}*x)}\right]_0^{\var{b4}}\right)\\ &=&\simplify[std]{({b4 ^ 2} / {n4}) * Exp({p}) -(({2 * b4} / {n4 ^ 2}) * Exp({p})) + (2 / {n4 ^ 3}) * (Exp({p}) -1)}\\ &=&\var{ans4}\mbox{ to 4 decimal places} \end{eqnarray*} \]

Question 2

Evaluate the following definite integrals.

a)

\[I=\int_0^{\var{b1}}\simplify[std]{e^({a}x)}\;dx\]

$I=\;\;$Expected answer:

Input your answer to 3 decimal places.

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b)

\[I=\int_0^{\var{b2}}\simplify[std]{1/({b}x+{m2})}\;dx\]

$I=\;\;$Expected answer:

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c)

\[I=\int_0^{\pi/2}\simplify[std]{({w} * Sin({m3} * x) + {1 -w} * Cos({m3} * x))}\;dx\]

$I=\;\;$Expected answer:

Input your answer to 3 decimal places.

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b)
\[\begin{eqnarray*}I&=&\int_0^{\var{b2}}\simplify[std]{1/(x+{m2})}\;dx\\ &=&\left[\ln(x+\var{m2})\right]_0^{\var{b2}}\\ &=& \ln(\var{b2+m2})-\ln(\var{m2})\\ &=&\ln\left(\frac{\var{b2+m2}}{\var{m2}}\right)\\ &=&\var{ans2}\mbox{ to 2 decimal places} \end{eqnarray*} \]

Question 3

Find the area $A$ of the shape bounded by the $x$-axis, the function $y=\simplify[std]{{t}*exp({b}/{c}*x)+{1-t}*ln({b}x+{c})}$ and the lines $x=\var{b1},\;x=\var{b2}$.

Enter the area $A$ here to 3 decimal places:

Expected answer:

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First we observe that:\[\simplify[std]{int ({t}*exp({b}/{c}*x)+{1-t}*ln({b}x+{c}),x)={t*c}/{b}*exp({b}/{c}*x)+({1-t}/{b}*({b}x+{c})*(ln({b}x+{c})-1))+C}.\]

Hence we have:

\[\begin{eqnarray*} \simplify[std]{defint ({t}*exp({b}/{c}*x)+{1-t}*ln({b}x+{c}),x,{b1},{b2})}&=&\left[\simplify[std]{{t*c}/{b}*exp({b}/{c}*x)+({1-t}/{b}*({b}x+{c})*(ln({b}x+{c})-1))}\right]_{\var{b1}}^{\var{b2}}\\&=&\var{ans}\end{eqnarray*}\]

to 3 decimal places.

Question 4

Consider the solid object that is obtained when the function: \[y=\simplify[std]{{a}(cos(x)+{b})}\] is rotated by $2\pi$ radians about the $x$-axis between the limits $x=\var{c}\pi$ and $x=\var{c+2}\pi$

Find the volume of this object.

$V=\;\;$Expected answer:

Enter your answer to 3 decimal places.

Click on Show steps for information on volumes of revolution. You will not lose any marks.

Recall that if $V$ is the volume generated between the limits $x=a$ and $x=b$ then $\displaystyle V=\pi\int_a^by^2\;dx$.

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Recall that if $V$ is the volume generated between the limits $x=a$ and $x=b$ by rotating the function about the $x$-axis then $\displaystyle V=\pi\int_a^by^2\;dx$.

So we have:
\[\begin{eqnarray*} V&=&\pi\int_{\var{c}\pi}^{\var{c+2}\pi}\simplify[std]{{a^2}(cos(x)+{b})^2}\;dx\\ &=&\var{a^2}\pi\int_{\var{c}\pi}^{\var{c+2}\pi}\simplify[std]{cos(x)^2+{2*b}*cos(x)+{b^2}}\;dx\\ &=&\var{a^2}\pi\left[\simplify[std]{((1 / 4) Sin(2*x) + (1 / 2) * x + {2 * b} * Sin(x) + {b ^ 2} * x)}\right]_{\var{c}\pi}^{\var{c+2}\pi}\\ \end{eqnarray*}\]
Here we have used the identity $\cos(x)^2=\frac{1}{2}(1+\cos(2x))$ in order to integrate $\cos(x)^2$.

Since $\sin(n\pi)=0$ for all integers $n$ we see that:
\[\begin{eqnarray*} V&=&\var{a^2}\pi\frac{\var{1+2b^2}}{\var{2}}\left(\var{c+2}\pi-\var{c}\pi\right)\\ &=&\var{a^2*(1+2b^2)}\pi^2\\ &=&\var{V}\mbox{ to 3 decimal places} \end{eqnarray*} \]

Question 5

Consider the function \[y=f(x)=\var{a1}\ln(\var{m}x)\]

a)

Rewrite this function in the form $x=g(y)$ , where $g(y)$ is a function of $y$ only.

$x=g(y)=\;\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\frac{ 1 }{ 6 } e^{ \frac{ y }{ 8 } }}$

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b)

Hence, find the volume of revolution, $V$ obtained as follows:

Rotate $y=f(x)$ by $2\pi$ radians about the $y$-axis, between the limits of $y=\var{c1}$ and $y=\var{d1}$.

The volume $V$ can be written as a multiple of $\pi$, $V=m\pi$, where:

$m=\;\;$Expected answer:. Input $m$ to two decimal places.

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a)
Given $y=\var{a1}\ln(\var{m}x)$ we rearrange so that $x$ is the subject of the equation:
\[\begin{eqnarray*} y&=&\var{a1}\ln(\var{m}x)\\ \Rightarrow \ln(\var{m}x) &=&\frac{y}{\var{a1}}\\ \Rightarrow \var{m}x&=& e^{\frac{y}{\var{a1}}}\\ \Rightarrow x&=&\frac{1}{\var{m}}e^{\frac{y}{\var{a1}}} \end{eqnarray*} \]
Hence \[g(y)=\frac{1}{\var{m}}e^{\frac{y}{\var{a1}}}\]

b)
The volume of revolution is given by:
\[V=\pi\int_{\var{c1}}^{\var{d1}}g(y)^2\;dy\]
Using the expression for $g(y)$ from the first part we have:
\[\begin{eqnarray*} V&=&\pi\int_{\var{c1}}^{\var{d1}}\frac{1}{\var{m^2}}\simplify[std]{e^(2y/{a1})}\;dy\\ &=&\simplify[std]{({a1}/{2*m^2})*pi}\left[\simplify[std]{e^(2y/{a1})}\right]_{\var{c1}}^{\var{d1}}\\ \\&=&\var{tvol}\pi = \var{vol}\pi \end{eqnarray*} \]
Hence the multiple of $\pi$ to two decimal places is $\var{vol}$.

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