Question

Find the area between the curve of y = sin0 and the 0 axis from `0 = pi/2 to 0 = 3pi/2` ?

Find the area between the curve of y = sin

Answer 1

0

Explanation:

Assuming it 0 was `theta`, we have:
Area beneath the curve of `y=sintheta` from `theta=pi/2` to `theta=3 pi/2`

We could rewrite this as: `int_(pi/2)^(3 pi/2)(sinx)dx`

To calculate this, we use the Fundemental Theorem of Calculs, which states that: `int_(a)^(b)f(x)dx=F(b)-F(a)` where `F'(x)=f(x)`

We calculate `intsinxdx`
Now, this is something you can remember.

`intsinxdx=-cosx`

We have:

`int_(pi/2)^(3 pi/2)(sinx)dx=-cos(3 pi/2)-(-cos(pi/2))`
=>`0+0`

=>`0`

Therefore,

`int_(pi/2)^(3 pi/2)(sinx)dx=0`

This is because we are counting the area above the `x` axis positive and the area below the `x` axis negative.