# 37. Find the area if the region bounded by the curves y = x^2 - 3x + 1 and y = x + 1. 38. Find...

## Question:

37. Find the area if the region bounded by the curves {eq}y = x^2 - 3x + 1 {/eq} and {eq}y = x + 1 {/eq}.

38. Find the area if the region bounded by the curves {eq}y = 2x^2 + x {/eq} and {eq}y = x^2 + 2 {/eq} from x = 0 to x = 2.

39. Find the function f(x) for which {eq}f'(x) = (x - 5)^2, \ f(8) = 2. {/eq}

40. Find the function f(x) for which {eq}f'(x) = {e}^{-5x}, \ f(0) = 1. {/eq}

## Area Using Calculus

Calculus can be used to find the area {eq}A {/eq} between the graphs of two function {eq}y=f(x) {/eq} and {eq}y=g(x) {/eq}. The method is to first find where the graphs intersect by solving {eq}f(x)=g(x) {/eq} for {eq}x {/eq}. Then determine which graph {eq}y=f(x) {/eq} or {eq}y=g(x) {/eq} is above the other graph on one or both sides of a point of intersection. After determining which graph is above the other the area can be calculated using calculus. If, for example, the graph {eq}y=f(x) {/eq} is above the graph {eq}y=g(x) {/eq} between {eq}x=a {/eq} and {eq}x=b {/eq} then the area {eq}A {/eq} is calculated as {eq}A=\int_{x=a}^b (f(x)-g(x))~dx {/eq}. The ideas just mentioned will be used to find the area {eq}A {/eq} for items 37 and 38. Items 39 and 40 involve solving a differential equation with an initial condition.

## Answer and Explanation:

The question is restated with slightly different notation.

37. Find the area if the region bounded by the curves {eq}y=x^2-3x+1 {/eq} and {eq}y=x+1 {/eq}.

38. Find the area if the region bounded by the curves {eq}y=2x^2+x {/eq} and {eq}y=x^2+2 {/eq} from {eq}x=0 {/eq} to {eq}x=2 {/eq}.

39. Find the function {eq}y=f(x) {/eq} for which {eq}\frac{dy}{dx}=f'(x)=(x-5)^2 {/eq}, {eq}f(8)=2 {/eq}.

40. Find the function {eq}y=f(x) {/eq} for which {eq}\frac{dy}{dx}=f'(x)={e}^{-5x} {/eq}, {eq}f(0)=1 {/eq}.

Consider item 37. Determine where the curves intersect by solving

{eq}\begin{eqnarray*} x+1 &=& x^2-3x+1 \\ 0 &=& x^2-4x \\ 0 &=& x(x-4). \end{eqnarray*} {/eq}

The solutions are {eq}x=0 {/eq} and {eq}x=4 {/eq}. The points of intersection are {eq}(0,1) {/eq} and {eq}(4,5) {/eq}. Next, determine which graph is above the other by picking a value of {eq}x {/eq} between 0 and 4, say {eq}x=1 {/eq}. We get

{eq}\begin{eqnarray*} y &=& 1+1 \\ &=& 2 \\ y &=& 1^2-3(1)+1 \\ &=& -1. \end{eqnarray*} {/eq}

Consequently, the curve {eq}y=x+1 {/eq} is above {eq}y=x^2-3x+1 {/eq} in the interval {eq}\left[0,4\right] {/eq}. Hence, the area {eq}A {/eq} bounded by the curves is

{eq}\begin{eqnarray*} A &=& \int_{x=0}^4 (x+1-(x^2-3x+1))~dx \\ &=& \int_{x=0}^4 (4x-x^2)~dx \\ &=& \left.2x^2-\frac{x^3}{3}\right|_{x=0}^4 \\ &=& 32-\frac{64}{3} \\ &=& \frac{96-64}{3} \\ &=& \frac{32}{3}. \end{eqnarray*} {/eq}

Consider item 38. Determine where the curves intersect by solving

{eq}\begin{eqnarray*} 2x^2+x &=& x^2+2 \\ x^2+x-2 &=& 0 \\ (x+2)(x-1) &=& 0. \end{eqnarray*} {/eq}

The solutions are {eq}x=-2 {/eq} and {eq}x=1 {/eq}. Only the value {eq}x=1 {/eq} is in the interval {eq}\left[0,2\right] {/eq} where we want to find the area. The point of intersection is {eq}(1,3) {/eq}. Next, determine which graph is above the other by picking a value of {eq}x {/eq} between 0 and 1, say {eq}x=0.5 {/eq}, and by picking a value of {eq}x {/eq} between 1 and 2, say {eq}x=1.5 {/eq}. In the interval {eq}\left[0,1\right] {/eq} we have

{eq}\begin{eqnarray*} y &=& 2(0.5)^2+0.5 \\ &=& 1.00 \\ y &=& (0.5)^2+2 \\ &=& 2.25. \end{eqnarray*} {/eq}

Consequently, the curve {eq}y=x^2+2 {/eq} is above {eq}y=2x^2+x {/eq} in the interval {eq}\left[0,1\right] {/eq}. In the interval {eq}\left[1,2\right] {/eq} we have

{eq}\begin{eqnarray*} y &=& 2(1.5)^2+1.5 \\ &=& 6.00 \\ y &=& (1.5)^2+2 \\ &=& 4.25. \end{eqnarray*} {/eq}

Consequently, the curve {eq}y=x^2+2 {/eq} is below {eq}y=2x^2+x {/eq} in the interval {eq}\left[1,2\right] {/eq}.

Hence, the area {eq}A {/eq} bounded by the curves is

{eq}\begin{eqnarray*} A &=& \int_{x=0}^1 (x^2+2-(2x^2+x))~dx + \int_{x=1}^2 (2x^2+x-(x^2+2))~dx \\ &=& \int_{x=0}^1 (-x^2+2-x)~dx + \int_{x=1}^2 (x^2+x-2)~dx \\ &=& \left.-\frac{x^3}{3}+2x-\frac{x^2}{2}\right|_{x=0}^1 + \left.\frac{x^3}{3}+\frac{x^2}{2}-2x\right|_{x=1}^2 \\ &=& \frac{7}{6}+\frac{11}{6} \\ &=& 3. \end{eqnarray*} {/eq}

Consider item 39. Solve the differential equation

{eq}\begin{eqnarray*} \frac{dy}{dx} &=& f'(x) \\ &=& (x-5)^2 \end{eqnarray*} {/eq}

subject to the boundary condition {eq}f(8)=2 {/eq}. Consider

{eq}\begin{eqnarray*} \frac{dy}{dx} &=& (x-5)^2 \\ dy &=& (x-5)^2dx \\ \int dy &=& \int (x-5)^2dx \\ y &=& \frac{(x-5)^3}{3} + C \\ f(x) &=& \frac{(x-5)^3}{3} + C. \end{eqnarray*} {/eq}

And so

{eq}\begin{eqnarray*} 2 &=& f(8) \\ 2 &=& \frac{(8-5)^3}{3} + C \\ 2 &=& \frac{27}{3}+C \\ 2 &=& 9+C \\ -7 &=& C. \end{eqnarray*} {/eq}

Therefore, the solution is {eq}y=f(x)=\frac{(x-5)^3}{3}-7 {/eq}.

Consider item 40. Solve the differential equation

{eq}\begin{eqnarray*} \frac{dy}{dx} &=& f'(x) \\ &=& e^{-5x} \end{eqnarray*} {/eq}

subject to the boundary condition {eq}f(0)=1 {/eq}. Consider

{eq}\begin{eqnarray*} \frac{dy}{dx} &=& e^{-5x} \\ dy &=& e^{-5x}dx \\ \int dy &=& \int e^{-5x}dx \\ y &=& -\frac{1}{5}e^{-5x} + C \\ f(x) &=& -\frac{1}{5}e^{-5x} + C. \end{eqnarray*} {/eq}

And so

{eq}\begin{eqnarray*} 1 &=& f(0) \\ 1 &=& -\frac{1}{5}e^{-5(0)} + C \\ 1 &=& -\frac{1}{5}+C \\ 1+\frac{1}{5} &=& C \\ \frac{6}{5} &=& C. \end{eqnarray*} {/eq}

Therefore, the solution is {eq}y=f(x)=-\frac{1}{5}e^{-5x}+\frac{6}{5} {/eq}.

#### Learn more about this topic:

Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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