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(a) The integral is equal to the area of the triangle; it has height 20 and base 10, so the area is 20*10/2 = 100.
(b) The integral is equal to the area of the semicircle with radius 10. It's also under the horizontal axis, so the area is negative. The semicircle has area , so the integral is -50π.
(c) First compute
which is the area of the triangle on the right. It has height and base 5, so its area is 25/2.
Then split up the desired integral as
and plug in the integral values you know:
Answer:
We have function,
Standard Form of Sinusoid is
Which corresponds to
where a is the amplitude
2pi/b is the period
c is phase shift
d is vertical shift or midline.
In the equation equation, we must factor out 2 so we get
Also remeber a and b is always positive
So now let answer the questions.
a. The period is
So the period is pi radians.
b. Amplitude is
Amplitude is 6.
c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.
So using that<em> rule, our range is [6+3, -6+3]= [9,-3] So our range</em> is [-3,9].
D. Plug in 0 for x.
So the y intercept is (0,-3)
E. To find phase shift, set x-c=0 to solve for phase shift.
Negative means to the left, so the phase shift is pi/4 units to the left.
f. Period is PI, so use interval [0,2pi].
Look at the graph above,
