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# Calculus II: Complete Set of Lessons

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L1-area.mws

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}); assign(ac);

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

>

>

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