// Numbas version: exam_results_page_options {"name": "Integration - Area Under a Graph 2", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"preventleave": false, "allowregen": true, "showfrontpage": false}, "question_groups": [{"questions": [{"variablesTest": {"maxRuns": 100, "condition": "AND(ymax2<100, 10(i) Area $= -\\int_{\\var{x11}}^0 \\simplify{{a1}*x^2+{b1}*x+{c1}} \\, dx +\\int^{\\var{x12}}_0 \\simplify{{a1}*x^2+{b1}*x+{c1}} \\, dx$

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

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

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

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

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

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

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(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:

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

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

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

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

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

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Now we can set-up the integral:

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

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

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

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Hence, using the integration by parts formula we get:

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

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

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

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

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

", "tags": [], "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}], "extensions": ["jsxgraph"], "statement": "

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

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

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

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

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[[0]]

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

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

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

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[[1]]

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

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

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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'.

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[[2]]

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