Definite integration 1

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Bill Foster

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Bill Foster 55 years, 2 months ago

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Name	Status	Author	Last Modified
Definite integration 1	draft	Bill Foster	06/08/2020 08:41
Julie's copy of Definite integration 1	draft	Julie Crowley	08/10/2016 20:51
cormac's copy of Julie's copy of Definite integration 1	draft	cormac breen	23/02/2017 18:19
Simon's copy of Julie's copy of Definite integration 1	draft	Simon Thomas	21/03/2019 09:27
Definite integration 1	draft	Jo-Ann Lyons	24/04/2019 15:47
Danny's copy of Definite integration 1	draft	Danny Tran	21/05/2019 05:12

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Question statement

Evaluate the following definite integrals.

Give any introductory information the student needs.

			Name	Type	Generated Value

ans1	number	82895.83
ans2	decimal	dec("0.63")
ans3	number	-1.57
ans4	number	0.0736
b4	integer	3
b1	integer	6
b2	integer	7
d1	integer	2
s2	integer	-1
s7	integer	1
s6	integer	-1
m4	integer	2
m3	integer	2
m2	integer	8
tol	number	0.01
a1	integer	7
tans4	number	0.07361243
c1	integer	1
tans1	number	82895.8333333333
tans3	number	-1.5707963268
tol1	number	0.0001
p	integer	-9
t	integer	1
w	integer	1
n4	integer	-3

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xxxxxxxxxx
 
precround(tans1,2)

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Generated value: `number`

82895.83

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tans1

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

$I=\;\;$ Unnamed gap

Input your answer to 2 decimal places.

Advice

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

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

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

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

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

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

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

Use integration by parts twice with $u=x^2$ and $\displaystyle \frac{dv}{dx}=\simplify[std]{e^({n4}x)}\Rightarrow v = \simplify[std]{1/{n4}e^({n4}x)}$

$\begin{eqnarray*} I&=&\left[\simplify[std]{x^2/{n4}Exp({n4} * x)}\right]_0^{\var{b4}}+\simplify[std]{2/{abs(n4)}DefInt(x*Exp({n4} * x),x,0,{b4})}\\ &=&\simplify[std]{{b4^2}/{n4}*e^{p}-2/{n4}}\left(\left[\simplify[std]{x/{n4}*e^({n4}*x)}\right]_0^{\var{b4}}+\simplify[std]{1/{abs(n4)}DefInt(e^({n4}x),x,0,{b4})}\right)\\ &=&\simplify[std]{{b4^2}/{n4}*e^{p}-2/{n4}}\left(\simplify[std]{{b4}/{n4}*e^{p}-1/{n4}}\left[\simplify[std]{1/{n4}*e^({n4}*x)}\right]_0^{\var{b4}}\right)\\ &=&\simplify[std]{({b4 ^ 2} / {n4}) * Exp({p}) -(({2 * b4} / {n4 ^ 2}) * Exp({p})) + (2 / {n4 ^ 3}) * (Exp({p}) -1)}\\ &=&\var{ans4}\mbox{ to 4 decimal places} \end{eqnarray*}$

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Bill Foster 55 years, 2 months ago

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Definite integration 1

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Taxonomy: mathcentre

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Bill Foster 55 years, 2 months ago

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Generated value: number

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