Calculus - final next week please help

Question

Q1 : Farmer Jones, and his wife, Dr. Jones, decide to build a fence in their field, to keep the sheep safe. Since Dr. Jones is a mathematician, she suggests building fences described by `y = 11 x^2 ` and `y = x^2 + 12 `. Farmer Jones thinks this would be much harder than just building an enclosure with straight sides, but he wants to please his wife. What is the area of the enclosed region?

Question 2. Sketch the region enclosed by ` y=4 x ` and ` y=8 x^2 `. Find the area of the region.

Answer 1

It is best first to always go to actually graphing these functions to see how they plot and the next steps in the integration.

When graphed, it is seen that x² + 12 is above 11 x² (as one could intuit when you see a ' + 12 ' term on one and not the other, but it is always good practice to double-check). When graphed, it is also seen that there are 2 intersection points. These points can be found by setting the 2 equations equal to each other and solving for x (intersections occur when functions equal one another).

x² + 12 = 11 x² 10 x² = 12 x² = 1.2 x ≈ ±1.1 (there's a lot of rounding going on for what x equals, I recommend you evaluate the square-root of 1.2 to see how irrational it is)

One thing to notice about these two quadratics is that they are symmetrical about the y-axis. You'd visually see it looking at the graph that plugging in 4 or - 4 gets you to the same height just at their specific x-position. You can see this mathematically because the intersection points did not have differing x-values; if the graphs were not symmetric about the y-axis, then the solution for intersections could have been x = -2 and x = 5.

That whole rant about graphical symmetry will come in handy with the integral to make it easier.

Next, because x² + 12 is graphically "higher" than 11 x² (this is to say, at every x-value between -1.1 to 1.1, x² + 12 has larger y-values than 11 x²), than your integral will take on this appearance:

^1.1∫ _{- 1.1} [ ( x² + 12 ) - ( 11 x² ) ] dx

And with a little bit of simplification:

^1.1∫ _{- 1.1} ( 12 - 10 x² ) dx

Now you could go ahead and evaluate that integral (no illegal moves there), but you can do one thing more to make it easier on yourself. When you have even functions (like cos(x), x², and many others), if your bounds are a reflection about the y-axis (the negative of one another), the integral across the original bounds is just twice the integral from zero to the positive bound. Said more explicitly:

^a∫ _{- a} f(x) dx = 2 * ^a∫ ₀ f(x) dx when f(x) is an even function ( f(-x) = f(+x) ; essentially intertwined with being symmetric about the y-axis )

So, the integal rewrites to:

^1.1∫ _{- 1.1} ( 12 - 10 x² ) dx = 2 * ^1.1∫ ₀ ( 12 - 10 x² ) dx = 2 * [ 12x - 10 / 3 x3 ] |₀^1.1

= 2 * [ (12 ( 1.1 ) - 10 / 3 * ( 1.1 )³ ) - ( 12 * (0) - 10 / 3 * ( 0 )³ ) ]

^1.1∫ _{- 1.1} ( 12 - 10 x² ) dx ≈ 17.53 (do NOT forget to double the integral from 0 to 1.1 at the end of calculating the integral alone; I almost forgot to in posting this)

The next integral is much simpler. Again, graphs are nice to look at to know how to approach your integral.

Find the intersection points like:

8 x² = 4x 2 x² - x = 0 x * ( 2x - 1) = 0 x = 0 2x - 1 = 0 x = 1 / 2

So the bounds of this integration are from 0 to 1 / 2.

4x is above 8 x² so the integral will take on this form:

^0.5∫ ₀ ( 4x - 8x² ) dx = ( 2 x² - 8 / 3 * x³ ) |₀^0.5 = ( 2 ( 0.5 )² - 8 / 3 * (0.5)³ ) - ( 8 / 3 * ( 0 )³ - 2 ( 0 )² )

^0.5∫ ₀ ( 4x - 8x² ) dx = 1 / 6 ≈ 0.1667

I hope this helps!

Answer 2

In each case you should graph the expressions first.

Then you need to find the intersection of the expressions by setting them equal to each other and solving for x.

Then integrate the difference of the two expressions between the points of intersection.

Messy, but not difficult.