Don't know how to solve it

Question

Find area between curve

y=3/2sin(pi(x)/6), y=3/6x

Answer 1

Following up on Mark M.'s insight that the second curve must be y = (3/6)x = x/2. (Prior to this, I just passed on the question.)

The curves y = x/2 and y = (3/2)sin(πx/6) cross three times: x = –3, x = 0, and x = 3.

(3/2)sin(πx/6) < x/2 on (–3, 0) and (3/2)sin(πx/6) > x/2 on (0, 3)

and the area between the two curves from x = –3 to x = 3

is twice the area you get integrating (3/2)sin(πx/6) – x/2 from x = 0 to x = 3:

A =

Answer 2

I'm going to assume you mean y = (3/6)x (which simplifies to y = x/2), because the graphs of y = (3/2)sin(πx/6) and y = x/2 cross each other at exactly two points (namely, the points (0, 0) and (3, 3/2)).

The graphs of y = (3/2)sin(πx/6) and y = 3/(6x) cross each other at infinitely many points. They wouldn't give you a problem like this without specifying which pair of these infinitely many points are the ones you should use as the left and right limits of integration.

So let's find the points where the graphs of y = (3/2)sin(πx/6) and y = x/2 cross each other. I simply drew the graphs of these two functions and guessed.

If you plug in x=0, you will find that (3/2)sin(πx/6) and x/2 are both the same number (namely, 0).

if you plug in x=3, you will find that (3/2)sin(πx/6) and x/2 are both the same number (namely, 3/2).

So the limits of integration are x=0 (on the left) and x=3 (on the right).

In the interval [0, 3], which function is greater?

Let's pick an x in this interval and evaluate both functions at that point.

I'll pick x=1, because when x=1, we have y=(3/2)sin(πx/6) = (3/2)sin(π/6)

π/6 radians is 30 degrees (because π radians is 180 degrees), and the sine of 30 degrees is 1/2.

(Picking x=1 got me an angle, 30 degrees, that I happened to know the sine of.)

So y=(3/2)sin(πx/6) = (3/2)(1/2) = 3/4 when x=1.

Meanwhile, the other curve y=x/2 is 1/2 when x=1.

So the curve y=(3/2)sin(πx/6) is above the curve y=x/2 when x=1.

Therefore the curve y=(3/2)sin(πx/6) is above the curve y=x/2 for the whole interval from x=0 to x=3,

because the two curves do not cross each other inside this interval.

We need to know which curve is upper and which is lower because the area between the two curves is

∫ (from x=0 to x=3) upper curve - lower curve dx

= ∫ (from x=0 to x=3) (3/2)sin(πx/6) - x/2 dx

Now let's integrate this integral. Remember, the derivative of cosine is negative sine, which means that the derivative of negative cosine is sine. We get

(3/2)(-cos(πx/6))(π/6) - (1/4)x² plus some constant, which simplifies to (-9/π)cos(πx/6) - (1/4)x².

Plugging in the left and right limits of integration, x=0 and x=3, we have

((-9/π)cos(π3/6) - (1/4)3²) - ((-9/π)cos(π0/6) - (1/4)0²)

= ((-9/π)cos(π3/6) - 9/4) - ((-9/π)cos(π0/6) - 0)

= ((-9/π)0 - 9/4) - ((-9/π)1 - 0)

= 9/π - 9/4

which is the area of the region between the two curves from x=0 to x=3.