• Exercise 1
• Exercise 2
• Exercise 3
• Exercise 4
• Exercise 5

Calculus II

Lesson 1: Area Between Curves

Exercise 1

and

(a) Plot both functions on the same axes. (b) Find the area of the region enclosed between the curves from x = 0 to x = 6.

> restart:

> with(plots):

Warning, the name changecoords has been redefined

> g:= x -> x^2 + 2;

> f:= x -> 2*x + 5;

> a:= plot(g(x), x = -1..7, thickness=2, color = red):

> b:= plot(f(x), x = -1..7, thickness=2, color = brown):

> display({a,b});

We need to find the intersection points for these two plots.

> solve(x^2 + 2 - 2*x - 5, x);

The area enclosed by these curves from x = 0 to x = 5 is

+

> int(2*x + 5 - x^2 - 2, x = 0..3);

> int(x^2 + 2 - 2*x - 5, x = 3..6);

Thus the total are enclosed is 27 + 9 = 36 sq units.

Let's redo the plot where we shade the areas in question.

> m:= plot(f(x), x = 0..7, thickness=2, color = magenta, axes=boxed):

> n:= plot(g(x), x = 0..7, thickness=2, color = blue,axes=boxed):

> p:= seq( plot([0 + i * (3/100) , t, t = g(0 + i*(3/100))..f(0 + i * (3/100))], thickness=2, color=red), i = 0..99):

> q:= seq( plot([3 + i * (3/100), t, t = f(3 + i *(3/100))..g(3 + i *(3/100))], thickness=2, color = brown), i = 1..100):

> r:= textplot([6,45,`g`], align=RIGHT):

> s:= textplot([6.5,7,`f`],align=RIGHT):

> display({m,n,p,q,r,s});

Exercise 2

The parabola is tangent to the graph of at two points and the area of the region bounded by their graphs is 10. Find a, b, and c. Make a sketch.

Solution :

The axis of the parabola is . That is also the axis of , so , or . The point where the slope of the parabola is 1 is on both graphs. Call the point [ ]. Then and .

> restart;

> eq1 := x0-1 = a*x0^2 -6*a*x0 + c;

> eq2 := 1 = 2*a*x0 - 6*a;

> ac := solve({eq1,eq2},{a,c});

Finally, the area between the curves is 100, so the righthand half is 50.

> eq3 := Int (a*x^2-6*a*x+c-(2+x-3),x=3..x0)=50;

> eq3 := int (a*x^2-6*a*x+c-(2+x-3),x=3..x0)=50;

> sol :=solve(subs(ac,eq3),x0);

> assign({x0=sol[1]}); assign(ac);

> plot([2+abs(x-3),a*x^2-6*a*x+c],x=sol[2]-2..sol[1]+2,thickness=2, color=[red,green], thickness=2);

Exercise 3

Sketch the region bounded by the given curves and find the area of the region.

x = 3y, x + y = 0, 7x + 3y = 24.

> with(plots):

Warning, the name changecoords has been redefined

> f:= x -> x/3;

> g:= x -> -x;

> h:= x -> (24 - 7*x) / 3;

> a:= plot(f(x), x = -.5..7, thickness=2, color = brown):

> b:= plot(g(x), x = -.5..7, thickness=2, color = blue):

> c:= plot(h(x), x = -.5..7, thickness=2, color = magenta):

> d:= textplot([6,4,`f`], thickness=2, color = brown):

> e:= textplot([1.5,7,`h`], thickness=2, color = magenta):

> k:= textplot([2,-4,`g`], thickness=2, color = blue):

> p:= seq( plot([0 + i * (3/50) , t, t = g(0 + i*(3/50))..f(0 + i * (3/50))], thickness=2, color=red), i = 1..50):

> q:= seq( plot([3 + i * (3/50), t, t = g(3 + i *(3/50))..h(3 + i *(3/50))], thickness=2, color = brown), i = 0..49):

> display({a,b,c,d,e,k,p,q});

We need to find the points of intersection.

We set f = g, h = g, and h = f.

> solve(f(x) = g(x), x);

> solve(h(x) = g(x), x);

> solve(h(x) = f(x), x);

Hence the desired area is given by

+

> int(x/3 + x, x= 0..3);

> int(8 - (7*x)/3 + x, x = 3..6);

Thus total area is 12 sq units.

Exercise 4

For what values of m do the line y = mx and the curve y =

enclose a region? Find the area of the region.

> restart:

> with(plots):

Warning, the name changecoords has been redefined

First lets plot the curve y =

along with an example of y = mx, where say m = .

.

> plot({x/2, x/(x^2 + 1)}, x = -.5...5, thickness=2);

We need to determine how large the slope m can be and still enclose

a region. From the above plot, the magnitude of m is determined

by the derivative of the curve y = at x = 0.

> f:= x -> x / (x^2 + 1);

> D(f);

> %(0);

Thus f ' (0) = 1. Hence a region is enclosed provided 0 < m < 1.

Let's replot the curve and shade the region enclosed for an example m, say m = 1/4.

Then we calculate the area for an arbitrary value of m.

> g:= x -> (1/4) * x;

> solve(f(x) = g(x), x);

> a:= plot(f(x), x = 0..2, thickness=2, color = blue):

> b:= plot(g(x), x = 0..2, thickness=2, color = brown):

> p:= seq( plot([0 + i * (sqrt(3)/50) , t, t = g(0 + i*(sqrt(3)/50))..f(0 + i * (sqrt(3)/50))], thickness=2, color=red), i = 1..49):

> display({a,b,p});

> solve( x / (x^2 + 1) = m * x,x);

Thus the intersection of the two curves is precisely when x = .

Hence the area is enclosed is given by:

.

Let's have maple do the integration.

> int(x / (x^2 + 1) - m*x, x = 0..sqrt(-m*(-1+m))/m);

Thus the area enclosed, for 0 < m < 1, is precisely:

.

Exercise 5

Find the value of d such that the area of the region bounded by the

parabolas y = - and y = -

is 576.

> restart:

> with(plots):

Warning, the name changecoords has been redefined

Let's get a sample picture. Say d = 3.

> f:= x -> x^2 - 9;

> g:= x -> 9 - x^2;

> solve(f(x) = g(x), x);

> a:= plot(f(x), x = -4..4, thickness=2, color = brown):

> b:= plot(g(x), x = -4..4, thickness=2, color = blue):

> c:= seq( plot([-3 + i * (6/50) , t, t = f(-3 + i*(6/50))..g(-3 + i * (6/50))], thickness=2, color=red), i = 1..49):

> display({a,b,c});

We can see from the above plot that in the general case the functions f and g intersect

at d and -d.

The enclosed area is given by

> int(2* d^2 - 2* x^2, x = -d..d );

> solve((8/3)*d^3 = 576, d)[1];

>

>

>

Thus when d = 6 or -6, the region enclosed has area 576.