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This is the second of our many quizzes on Vertical and horizontal slicing. Click on this to access more such competitive quizzes on Calculus.

Vertical and horizontal slicing Quiz 2

mathematics

    The function ${{\sin }^{2}}(x)$ is sliced vertically between $x=4$,and $x=8$. Calculate the area in between the two lines. 

 Use the reduction formula for ${{\sin }^{2}}(x)$ to simplify the integral. 

  $ f(x)={{\sin }^{2}}(x) \\   \int\limits_{4}^{8}{{{\sin }^{2}}(x)dx} \\   \int{{{\sin }^{2}}(x)dx=\dfrac{n-1}{n}\int{{{\sin }^{n-2}}(x)dx-\dfrac{\cos (x){{\sin }^{n-1}}(x)}{n}}} \\   -\dfrac{\cos (x)\sin (x)}{2}+\dfrac{x}{2}=-\dfrac{\sin (2x)-2x}{4} \\ \text{Evaluating } -\dfrac{\sin (2x)-2x}{4} \text{ from x=4 to x=8,}\\  \left( \dfrac{\sin (8)-8}{4}-\dfrac{\sin (16)-16}{4} \right)=2.32$

 For a function, ${{\cos }^{2}}x$, a teacher slices the function vertically. Given the interval $[0,2\pi ]$, calculate the area formed in the given interval. 

 Use the reduction formula for ${{\cos }^{2}}(x)$ to simplify the integral.

  $ \int\limits_{0}^{2\pi }{{{\cos }^{2}}(x)dx} \\   \int{{{\cos }^{n}}(x)dx=\dfrac{n-1}{n}\int{{{\cos }^{n-2}}(x)dx+\dfrac{{{\cos }^{n-1}}\sin (x)}{n}}} \\  \dfrac{\cos (x)\sin (x)}{2}+\dfrac{x}{2}=\dfrac{\cos (x)\sin (x)+x}{2} \\  \text{Evaluating } \dfrac{\cos (x)\sin (x)+x}{2} \text{ from x=0 to } x=2\pi ,\\ \left( \dfrac{\cos (2\pi )\sin (2\pi )+2\pi }{2} \right)-\left( \dfrac{\cos (0)\sin (0)+0}{2} \right)=\pi $

  A parabola $y=2{{x}^{2}}$ is rotated along the $x-$axis such that it creates a $3-$dimensional figure. The parabola is sliced vertically. Find the parametric equations of the curves in terms of ${{x}_{0}}$. 

 A set of parametric equations can be used when two variables are not related but are related to another third variable. Use ${{x}_{0}}$to find the set of parametric equations. 

 Assign the variable ${{x}_{0}}$ since $x$ and $y$ are not related directly but are related to a third variable. $ \\ $ $x={{x}_{0}}$ and $y=2{{x}_{0}}^{2}$

  Two parabolas are sliced vertically. Given that the function of one the parabola is ${{x}^{2}}+6x+9$ and the other has a value of $a=1$ for the quadratic function $a{{x}^{2}}+bx+c$ which makes it a perfect square. Calculate the possible equation and the area of the second parabola given that the two functions intersect at $y=16$ where $x\ge 0$.

  Calculate the $x-$value where both the parabolas intersect each other and create 2 separate integrals allowing you to calculate the area. 

 $  {{x}^{2}}+6x+9,{{x}^{2}}+bx+25 \\   {{x}^{2}}+6x+9=16 \\   {{x}^{2}}+6x=7 \\  (x+7)(x-1)=0 \\   x=1,-7 \\   {{x}^{2}}-bx+25,b=10 \\  {{x}^{2}}-10x+25 \\ \text{It is known that the vertex of } {{x}^{2}}+6x+9 \text{ is x=-3 } \\ \text{Whilst the vertex of } {{x}^{2}}-10x+25  \text{ is x=5 },\\ \text{therefore, }\int\limits_{-3}^{1}{{{x}^{2}}+6x+9}dx+\int\limits_{1}^{5}{{{x}^{2}}-10x+25}dx\\ \dfrac{{{x}^{3}}}{3}+3{{x}^{2}}+9x \text{ evaluated from x=-3 to x=1 } \text{ and} \\ \dfrac{{{x}^{3}}}{3}-5{{x}^{2}}+25x \text{ from x=1 to x=5} \\ \text{Hence, the area is } \dfrac{64}{3}+\dfrac{64}{3}=\dfrac{128}{3} $

 $\ln ({{x}^{e}})$ is a logarithmic function. Given that the area is calculated using vertical slices, find the Trapezoidal Riemann sum of the area enclosed in the interval $[1,10]$ by using $5$ slices.

 Use the trapezoidal rule which is $\int\limits_{a}^{b}{f(x)dx\approx \dfrac{\Delta x}{2}\left( f({{x}_{0}})+f({{x}_{1}})+f({{x}_{2}})+f({{x}_{n-1}})+f({{x}_{n}}) \right)}$ such that you can calculate the trapezoidal Riemann sum in the given interval of the function. 

$ \text{Using the trapezoidal rule,} \\  \int\limits_{a}^{b}{f(x)dx\approx \dfrac{\Delta x}{2}\left( f({{x}_{0}})+f({{x}_{1}})+f({{x}_{2}})+f({{x}_{n-1}})+f({{x}_{n}}) \right)} \\  \Delta x=\dfrac{b-a}{n}=\dfrac{10-1}{5}=\dfrac{9}{5} \\ \text{Dividing the interval } [1,10] \text{ into 10 subintervals with } \Delta x=\dfrac{9}{5}, \text{ then, the endpoints are:} \\ a=1,\dfrac{14}{5},\dfrac{23}{5},\dfrac{32}{5},\dfrac{41}{5},10=b \\   f({{x}_{0}})=f(a)=f(1)=0 \\   2f({{x}_{1}})=2f\left( \dfrac{14}{5} \right)=2\ln \left( {{\left( \dfrac{14}{5} \right)}^{e}} \right)=5.598 \\   2f({{x}_{2}})=2f\left( \dfrac{23}{5} \right)=2\ln \left( {{\left( \dfrac{23}{5} \right)}^{e}} \right)=8.297 \\   2f({{x}_{3}})=2f\left( \dfrac{32}{5} \right)=2\ln \left( {{\left( \dfrac{32}{5} \right)}^{e}} \right)=10.092 \\   2f({{x}_{4}})=2f\left( \dfrac{41}{5} \right)=2\ln \left( {{\left( \dfrac{41}{5} \right)}^{e}} \right)=11.439 \\   f({{x}_{5}})=f\left( b \right)=\ln \left( {{\left( 10 \right)}^{e}} \right)=6.259 \\ \dfrac{\Delta x}{2}=\dfrac{9}{10} \\   \dfrac{9}{10}\left( 0+5.597+8.297+10.902+6.259 \right)=37.5  $