Integration - Area Under a Graph 2 - Numbas at mathcentre.ac.uk

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Kevin Bohan 5 years, 7 months ago

Created this as a copy of Integration: calculating areas in graphs using definite integration..

Name	Status	Author	Last Modified
Integration: calculating areas in graphs using definite integration.	Ready to use	Lovkush Agarwal	06/08/2018 13:30
Algebra: solving equations, minimum and maximum $x$ -values of a region	Ready to use	Lovkush Agarwal	06/08/2018 13:34
solving equations, minimum and maximum $x$ -values of a region	draft	Xiaodan Leng	10/07/2019 23:37
Integration: calculating areas in graphs using definite integration.	draft	Xiaodan Leng	10/07/2019 23:38
Integration - Area Under a Graph 2	draft	Kevin Bohan	26/09/2019 09:26

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Question statement

For each graph, determine the integral that corresponds to the shaded areas.

Give any introductory information the student needs.

			Name	Type	Generated Value

a1	integer	4
b1	integer	5
c1	integer	0
x12	integer	5
x11	integer	-1
ymax1	integer	127
area1	decimal	dec("230")
a2	integer	1
b2	integer	3
c2	integer	3
x21	integer	-1
x22	integer	2
ymax2	number	20.0855369232
area2	number	7.61
a3	integer	5
b3	integer	2
x31	integer	0
x32	number	2 pi
area3	number	20

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Generated value: `integer`

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area1
ymax1

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2. Gap Gap 1. Number entry
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(i) {plotgraph(1,x11,x12,a1,b1,c1,{x11-2},{x12+1},-3*ymax1/10,ymax1*1.1)}

This is the graph of the function $f(x) = \simplify{{a1}*x^2+{b1}*x+{c1}}$.

What is the total area of the shaded region? (In case it is unclear, the minimum $x$-value of the region is {x11}).

Gap 0.

(ii) {plotgraph(2,x21,x22,a2,b2,c2,x21-1,x22+1,-3*ymax2/10,ymax2*1.1)}

This is the graph of the function $f(x) = \simplify{{a2}e^({b2}*x-3)+1/({c2}+x)}$.

Find the area of the shaded region. (In case it is unclear, the minimum $x$-value of the region is {x21}).

Gap 1.

(iii) {plotgraph(3,x31,x32,a3,b3,0,-1,x32+1,-4,x32*a3/1.4)}

This curve has equation $y = \simplify{{a3}*x*sin(x/{b3})}$.

Find the area of the shaded region. Provide the exact answer - it will be a multiple of $\pi$. To enter $\pi$ in the answer box, you type `pi'.

Gap 2.

Advice

(i) Area $ = -\int_{\var{x11}}^0 \simplify{{a1}*x^2+{b1}*x+{c1}} \, dx +\int^{\var{x12}}_0 \simplify{{a1}*x^2+{b1}*x+{c1}} \, dx$

$ = -\left[ \simplify{ {a1}/3*x^3 + {b1}/2*x^2 + {c1}*x } \right]_{\var{x11}}^0 + \left[\simplify{ {a1}/3*x^3 + {b1}/2*x^2 + {c1}*x } \right]^{\var{x12}}_0 $

$ =-\left[ (0) - (\simplify{{a1}/3{x11}^3 +{b1}/2{x11}^2 +{c1}{x11}} ) \right] + \left[(\simplify{{a1}/3({x12})^3 +{b1}/2({x12})^2 +{c1}{x12}}) -(0) \right] $

$ = \var{area1}$, to 3.s.f.

(ii) Area $ = \int_{\var{x21}}^{\var{x22}} \simplify{{a2}e^({b2}*x-3)+1/({c2}+x)} \, dx$

$ = \left[ \simplify{{a2}/{b2}*e^({b2}*x-3) + ln(x+{c2}) } \right]_{\var{x21}}^{\var{x22}}$

$ = \var{area2}$, to 3.s.f.

(iii) First we need to work out the minimum and maximum $x$-values. The minimum can be read from the graph, it is $0$. The maximum is found by solving an equation:

$\simplify{{a3}*x*sin(x/{b3})} =0$

$\sin(\frac{x}{\var{b3}}) = 0 $

$\simplify{x/{b3}} =\ldots,-2\pi, -\pi,0,\pi,2\pi,3\pi,\ldots$, (obtained by looking at the graph of $\sin(x)$)

$ x = \ldots, -\var{2*b3} \pi, -\var{b3}\pi, 0 , \var{b3}\pi,\var{2*b3}\pi,\var{2*b3}\pi, \ldots$. (These values were obtained by multiplying the previous line by $\var{b3}$.)

We need the smallest positive value, which is $\var{b3}\pi$.

Now we can set-up the integral:

Area $ = \int^{\var{b3}\pi}_0 \simplify{{a3}*x*sin(x/{b3})} \, dx$.

To integrate one can use integration by parts. First we let $u = x$ and $\frac{dv}{dx}=\simplify{{a3}sin(x/{b3})}$.

Differentiating $u$ gives $\frac{du}{dx} = 1$ and integrating gives $v = \simplify{-{a3*b3}cos(x/{b3})}$.

Hence, using the integration by parts formula we get:

Area $= \left[ x \times \simplify{-{a3*b3}cos(x/{b3})} \right]^{\var{b3}\pi}_0 - \int^{\var{b3}\pi}_0 1 \times \simplify{-{a3*b3}cos(x/{b3})} \, dx$

$ =\left[ \simplify{-{a3*b3}*x*cos(x/{b3})} \right]^{\var{b3}\pi}_0 + \left[ \simplify{{a3*b3*b3}sin(x/{b3}) }\right]^{\var{b3}\pi}_0 $

$ = [( -\var{a3*b3*b3}\pi \times \cos(\pi)) - (0) ] + [\var{a3*b3*b3}\sin(\pi) -\var{a3*b3*b3}\sin(0)]$

$ = [ -\var{b3*a3*b3}\pi \times -1] + [0 -0]$

$ = \var{a3*b3*b3} \pi$.

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Integration - Area Under a Graph 2

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Kevin Bohan 5 years, 7 months ago

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Integration - Area Under a Graph 2

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Taxonomy: mathcentre

Taxonomy: Kind of activity

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Kevin Bohan 5 years, 7 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`