Definite integration using standard integrals

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Newcastle University Mathematics and Statistics

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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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	433.97
ans2	decimal	dec("0.98")
ans3	number	1.05
ans4	number	0.6466
b4	integer	2
b1	integer	2
b2	integer	10
d1	integer	2
s2	integer	1
s7	integer	1
s6	integer	-1
m4	integer	2
m3	integer	3
m2	integer	6
tol	number	0.01
a1	integer	7
tans4	number	0.6466471676
c1	integer	1
tans1	number	433.9666666667
tans3	number	1.0471975512
tol1	number	0.0001
p	integer	-2
t	integer	1
w	integer	1
n4	integer	-1

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Generated value: `number`

433.97

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tans1

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

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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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This question is used in the following exams:

Integration using standard integrals by Newcastle University Mathematics and Statistics in Content created by Newcastle University.
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Definite integration using standard integrals

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Definite integration using standard integrals

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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