Area bounded by curves

Question

I'm having trouble solving these problems, can someone please go through this question step by step so I can do other questions similar to this?

Find the area bounded by the curves $y = x^2 - 6x + 7$ and $y = x-3$ . Thanks!

Answer 1

First calculate the intersection points of both curves, then calculate the integral of the first curve minus the second one, the integral should be taking from one intersection point to the other.

So we find that $x^2-6x+7=x-3$ if and only if $x=2$ or $x=5$. Notice that on the interval $[2,5]$, $x-3$ is above the other curve. Hence the enclosed area is

$\int_{2}^5 x-3-(x^2-6x+7)\mathrm{d}x$

Answer 2

You have to integrate both functions between the left $x_1$ and the right $x_2$ boundary. You didn't specify the boundaries but it may be in particular the intersection values if they exist. To find them, find the roots of the function $f'_3(x)=f'_2(x)-f'_1(x)=x^2-6x+7-(x-3)=x^2-7x+10$. Here both curves intersect only for x=2 and x=5.

if $f_1'(x)=x^2-6x+7$, then $f_1(x)=x^3/3-3x^2+7x+cst$ and its area between $x_1$ and $x_2$ is $f_1(x_2)-f_1(x_1)$

if $f_2'(x)=x-3$, then $f_2(x)=x^2/2-3x+cst$ and its area between $x_1$ and $x_2$ is $f_2(x_2)-f_2(x_1)$

Then substract the area of the lower curve to the area of the upper curve.

Answer 3

It would be great if you start by ploting the curves, so you can visualize the region your are seeking for its area. Simply you can use any online plotter, see for example FooPlot. For the specific case you give, we got this plot

Second, as it appears from that your curves have two intersection points in which your region is within. Try to find these intersection points taking $y=y$, the you will got the abscisas $x_0$ and $x_1$ of the intersection points.

The area then is calculated by computing the double integral of 1 over the region. In a general case you may take this double integral as $$ \int_{x_0}^{x_1} \int_{y_0}^{y_1} 1 dydx $$ where $x_0 \leq x_1$, and $y_0$ is the curve bounding your region from below, and $y_1$ the curve bounding your region from above. So for example in your case, your Area $A$ is $$ A=\int_{2}^{5} \int_{x^2-6x+7}^{x-3} dydx $$

Answer 4

A completely ad-hoc solution:

The area between the two curves is the same as the area between the $y$ axis and the "difference curve" $$y=(x-3)-(x^2-6x+7)=\dfrac94-\left(x-\dfrac72\right)^2.$$

This is a segment of a parabola, with width $3$ (the roots are at $x-\frac72=\pm\frac32$), and height $\frac94$.

By a known formula,

$$A=\frac23wh=\frac92.$$