Exam14 - Definite integration - Numbas at mathcentre.ac.uk

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Michael Proudman 5 years, 9 months ago

Created this as a copy of Definite integration using standard integrals.

Name	Status	Author	Last Modified
Definite integration using standard integrals	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Exam14 - Definite integration	draft	Michael Proudman	07/12/2021 09:54
Shivram's copy of Exam14 - Definite integration	draft	Shivram Venkat	09/06/2019 18:49
nuExam14 - Definite integration	draft	Michael Proudman	07/12/2021 09:54

There are 15 other versions that do you not have access to.

Question statement

Evaluate the following definite integrals.

Give any introductory information the student needs.

			Name	Type	Generated Value

ans1	number	62.27
ans2	decimal	dec("0.69")
ans3	number	-0.52
ans4	number	0.0695
b4	integer	2
b1	integer	3
b2	integer	7
d1	integer	-8
s2	integer	-1
s7	integer	1
s6	integer	-1
m4	integer	2
m3	integer	6
m2	integer	7
tol	number	0.01
a1	integer	3
tans4	number	0.0694837923
c1	integer	-4
tans1	number	62.2666666667
tans3	number	-0.5235987756
tol1	number	0.0001
p	integer	-6
t	integer	-1
w	integer	1
n4	integer	-3

Automatically regenerate variables?

Name

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xxxxxxxxxx
 
precround(tans1,2)

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Generated value: `number`

62.27

← Depends on:

tans1

→ Used by:

Advice
"Part a)" → "Unnamed gap" - Minimum accepted value
"Part a)" → "Unnamed gap" - Maximum accepted value

Parts

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Part b) Gap-fill
1. Gap Number entry
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Part c) Gap-fill
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Part d) 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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This question is used in the following exam:

Exam Practice by Michael Proudman in Assessment Exercises.

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Exam14 - Definite integration

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Michael Proudman 5 years, 9 months ago

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Exam14 - Definite integration

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Taxonomy: mathcentre

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Michael Proudman 5 years, 9 months ago

Error in variable testing condition

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Generated value: number

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Gap-fill

Generated value: `number`