Integration question?

Question

What is the area bounded by the function f(x)=sin(x) and the x-axis on the interval [0,3pi/2]?

Answer 1

∫sin(x) dx = -cos(x) The area in the interval from 0 to π is above the x-axis and is positive. The area from π to 3π/2 is under the x axis, and the integral will give a negative result. However, area is always positive, so the integration should be done in two sections:

∫[0 to π] sinx dx - ∫[π to 3π/2] sinx dx

-cosx [@π - @0] + cosx[@3π/2 - @π]

The result is 3.

Answer 2

If we differentiate cos(X) we get -sin(X), but we want to end up with sin(X) (no minus sign) so we must differentiate -cos(X). That means that the integral of sin(X) is -cos(X).
Remember that to find the area you follow the formula
F(b) - F(a)
where F(X) is the integral of the function f(X)
Now since you want the area from [0,3π/2], let b = 3π/2 and a = 0/
A = F(3π/2) - F(0)
= -cos(3π/2) - (-cos(0) )
= 0 - (-1)
= 0+1
=1

A = 1

Answer 3

3 units

Answer 4

3 times.integral from 0 to pi/2 of sinx

-cosx from 0 to pi/2
-cos(pi/2)+cos0
=-0+1
=1

area can not be negative

Answer 5

int sinx = -cosx, right? because d/dx (-cosx) = -(-sinx) = sinx.
so plug in the limits
-cos3pi/2 - -cos0 = 0 + 1 = 1.