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This is the second of our many quizzes on Area between curves. Click on this to access more such competitive quizzes on Calculus.

Area between curves Quiz 2

mathematics

 Satish wants to place a metal sheet in such a way that it’s most aerodynamic and is in the shape of equation $3-{{y}^{2}}$. A building is placed such that it is one unit in the opposite direction of the sheet, plotted using a negative sign on the graph. Find the area so formed in between the metal sheet and the building  

 Find the limits of the integral by keeping the value of $x=-1$ in the equation. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c10bf85b-b955-4a99-af18-85e047fbdb6a463124466872364921.png"/> $ \\ $ The equation $3-{{y}^{2}}$ is of a parabola facing towards the negative x axis, covering area with $y=-1$ $ \\ $  The limits are $-2\le x\le 2$ $ \\ $ Since $x=-1$ according to the equation in the question $ \\ 3+2x-{{x}^{2}}=0\to -(x+1)(x-3)=0\to x=-1,x=3 \\ $ Since the limits are known, evaluating the integral $ \\ Area=\int_{-2}^{2}{3-{{y}^{2}}-(-1)dy=\left[ 4y-\dfrac{1}{3}{{y}^{3}} \right]_{-2}^{2}=\dfrac{32}{3}}$

 Ram wants to use concepts of math with art, he creates a mural with equations $y=x\sqrt{{{x}^{2}}+1}$ , $y={{e}^{-\frac{1}{2}x}}$ and one side is a straight line as $x=-3$. The mural covers one side of the wall given as $y-axis$. What area does he cover in the mural? 

 The limits of the functions are given as the equations of two straight lines so $-3\le x\le 0$ Substitute these in the equation to integrate and find the solution.

  The limits of the functions are given as the equations of two straight lines so $-3\le x\le 0$   $ \\ Area=\int_{-3}^{0}{{{e}^{-\frac{1}{2}x}}}-(x\sqrt{{{x}^{2}}+1})dx \\ =\left[ -2{{e}^{-\frac{1}{2}x}}-\dfrac{1}{3}{{({{x}^{2}}+1)}^{\frac{3}{2}}} \right]_{-3}^{0} \\ = -\dfrac{7}{3}+2{{e}^{\frac{3}{2}}}+\dfrac{1}{3}{{10}^{\frac{3}{2}}}$

 A physicist wants to find the surface tension on the panel shaped by equations $y=4x+3$ and $y=6-x-2{{x}^{2}}$. For surface tension he needs to find the area first. If the other two sides are given by equations $x=-4$ and $x=2$. Help her by finding the area of the panel.

 Find the area of all 3 regions covered by the equations with limits as $-4\le x\le -3$ , $-3\le x\le \dfrac{1}{2}$ and $\dfrac{1}{2}\le x\le 2$. Substitute these in the equation to integrate and find the solution.

Calculating the intersection points of the equations to find the limits of the function $ \\ 6-x-2{{x}^{2}}=4x+3 \\ 2{{x}^{2}}+5x-3=(2x-1)(x+3)=0 \\ x=-3,x=\dfrac{1}{2} \\ $ Since we have $4$ values of $x$ we need to find the area between $3$ regions given by the limits $ \\ $ $-4\le x\le -3$ , $-3\le x\le \dfrac{1}{2}$ and $\dfrac{1}{2}\le x\le 2$ $ \\ $ Since the limits are known, integrating the expression $ \\A=\int_{-4}^{-3}{4x+3-(6-x-2{{x}^{2}})dx+}\int_{-3}^{\frac{1}{2}}{6-x-2{{x}^{2}}-(4x+3)dx+}\int_{\frac{1}{2}}^{2}{4x+3-(6-x-2{{x}^{2}})dx} \\ A=\left( \dfrac{2}{3}{{x}^{3}}+\dfrac{5}{2}{{x}^{2}}-3x \right)_{-4}^{-3}+\left( -\dfrac{2}{3}{{x}^{3}}-\dfrac{5}{2}{{x}^{2}}+3x \right)_{-3}^{\frac{1}{2}}+\left( \dfrac{2}{3}{{x}^{3}}+\dfrac{5}{2}{{x}^{2}}-3x \right)_{\frac{1}{2}}^{2} \\ A=\dfrac{25}{6}+\dfrac{343}{24}+\dfrac{81}{8}=\dfrac{343}{12}$ 

 Sahil wishes to make a new component for his machine, the component is shaped by equations $y=\dfrac{1}{x+2}$ and $y={{(x+2)}^{2}}$. If the other two sides are given by equations $x=\dfrac{-3}{2}$ and $x=1$. Find the metal required to make the component. 

 Calculating the intersection points of the equations to find the limits of the function  $ \\ \dfrac{1}{x+2}={{(x+2)}^{2}} \\ 1={{(x+2)}^{3}} \\ x=-1 \\ $ Since we have $3$ values of $x$ we need to find the area between $2$ regions given by the limits $ \\ $  $-1\le x\le -1$ , and $-\dfrac{3}{2}\le x\le -1$ $ \\ $ Since the limits are known, evaluating the integral $ \\ A=\int_{\frac{-3}{2}}^{-1}{\dfrac{1}{x+2}-{{(x+2)}^{2}}dx+}\int_{-1}^{1}{{{(x+2)}^{2}}-\dfrac{1}{x+2}dx} \\ A=\left( \ln \left| x+2 \right|-\dfrac{1}{3}{{(x+2)}^{3}} \right)_{-\frac{3}{2}}^{-1}+\left( \dfrac{1}{3}{{(x+2)}^{3}}-\ln \left| x+2 \right| \right)_{-3}^{\frac{1}{2}} \\ A=-\dfrac{7}{24}-\ln \left( \dfrac{1}{2} \right)+\dfrac{26}{3}-\ln \left( 3 \right)=\dfrac{67}{8}-\ln \left( \dfrac{1}{2} \right)-\ln \left( 3 \right)$ 

 Suppose you want to make a base for your science project with equations as $x={{y}^{2}}+1$ and $x=5$. If the other two sides are given by equations $y=-3$ and $y=3$. If the cost of the wood is Rs 3 per sq. unit. Find the cost of the base required.

 Since we have 4 values of y we need to find the area between 3 regions given by the limits$-3\le x\le -2$ , and $-2\le x\le 2$ and $2\le x\le 3$. Substitute these in the equation to integrate and find the solution.

 Finding the limits of the functions by calculating the intersection points of the equations. $ \\ {{y}^{2}}+1=5 \\ {{y}^{2}}=4 \\ y=+2,y=-2 \\ $ Since we have $4$ values of $y$ we need to find the area between $3$ regions given by the limits $ \\ -3\le x\le -2$ ,$-2\le x\le 2$$2\le x\le 3 \\ $ Since the limits are known, evaluating the integral $ \\ A=\int_{-3}^{-2}{{{y}^{2}}+1-5dy+}\int_{-2}^{2}{5-\left( {{y}^{2}}+1 \right)dy+}\int_{2}^{3}{\left( {{y}^{2}}+1 \right)-5dy} \\ A=\left( \dfrac{1}{3}{{(y)}^{3}}-4y \right)_{-3}^{-2}+\left( 4y-\dfrac{1}{3}{{(y)}^{3}} \right)_{-2}^{2}+\left( \dfrac{1}{3}{{(y)}^{3}}-4y \right)_{2}^{3} \\ A=\dfrac{7}{3}+\dfrac{32}{3}+\dfrac{7}{3}=\dfrac{46}{3} \\ $ Cost is the product of rate and area which is equal to 46.