1. Integrals and Graphing

I'm having some trouble with the following problems...

Sketch the region bounded by the graphs of the functions and find the area of the region:
1) y=x^3 - 2x + 1, y=-2x, x=1
2) y=(e^(1/x))/(x^2), y=0, x=1, x=3

Write the definite integrals that represent the area of the region:
3) y=4/x, y=x, x=1, x=4

4) Find the area of the region enclosed by the graphs of f(x)=sinx and g(x)=cosx between x=pi/4 and x=5pi/4.

I have absolutely no idea what to do for these probs, so any help would be much appreciated. Thanks

2. Originally Posted by cupcakelova87
I'm having some trouble with the following problems...

Sketch the region bounded by the graphs of the functions and find the area of the region:
1) y=x^3 - 2x + 1, y=-2x, x=1
okay, so i'm not at home and hence cannot use all the tools i usually use to make the graphs look pretty, by shading in the area that you want to find and such. so you have to look at the basic graph and figure it out. we want to find the area enclosed by the red, blue and green curves. the point of sketching was to figure out which graph is above which. why, because the area will be given by: int{higher graph - lower graph}dx

now we need to find our limits of integration. the last one is easy, it's x = 1, they told us that. the first one is where the graphs intersect, so let's find that by equating the functions.

y = x^3 - 2x + 1
y = -2x

=> for intersection: x^3 - 2x + 1 = -2x
=> x^3 + 1 = 0 ...............this is the sum of two cubes, there's a formula for that (the alternative is doing long division or synthetic division..yuck!)
=> (x + 1)(x^2 - x + 1) = 0
=> x = -1 since x^2 - x + 1 has no real roots.

so our limits of integration is from -1 to 1

so, Area = int{x^3 - 2x + 1 - (-2x)}dx
............= int{x^2 + 1}dx
............= (1/3)x^3 + x evaluated between -1 and 1, we use the fundamental theorem of calculus, i hope you know it
=> Area = (1/3)(1)^3 + 1 - (1/3)(-1)^3 - 1 = 2/3

3. Thanks Jhevon, you never fail to impress me I have a quick question though...why does x^3 become x^2? Is that a mistake or am I blind?
"so, Area = int{x^3 - 2x + 1 - (-2x)}dx
............ = int{x^2 + 1}dx"

I think I was able to figure out #3 and #4...still stuck on #2 though. Any way you can help me with that one?

4. Originally Posted by cupcakelova87
Thanks Jhevon, you never fail to impress me I have a quick question though...why does x^3 become x^2? Is that a mistake or am I blind?
"so, Area = int{x^3 - 2x + 1 - (-2x)}dx
............ = int{x^2 + 1}dx"
you are correct, the x^2 was a typo that i followed through on make the necessary corrections.

I think I was able to figure out #3 and #4...still stuck on #2 though. Any way you can help me with that one?
2) y=(e^(1/x))/(x^2), y=0, x=1, x=3

Ok, let's see where we get with this one. Draw the graph first, see below. So this graph would be hell to draw by hand, but just the way the question is set out we can know where to go even if we don't graph. first thing you should note is that this function is never zero, so we are always above the x-axis. they gave us the limits, so we don't have to figure those out, yay! now we are bounded by two vertical lines and the y-axis, so our integral is just going to be the integral of the function between the given limits.

Area = int{e^(1/x))/(x^2)}dx evaluated between x=1 to x = 3

we proceed by substitution
let u = 1/x
=> du = -1/x^2 dx
=> -du = 1/x^2 dx
so our integral becomes:
-int{e^u}du = -e^u + C = -e^(1/x) + C
now by the fundamental theorem of calculus, when we evaluate between the desired limits, we get:
Area = -e^(1/3) + e^(1/1) = e - e^(1/3)