Evaluating the integral of absolute values

[doctordiddy]

Homework Statement

∫(0 to 3pi/2) -7|sinx|dx

Homework Equations

The Attempt at a Solution

I am not sure how to treat it as it has an absolute value

i assumed that you could remove the -7 to get

-7∫|sinx| dx

then integrate sinx into -cosx but since there is absolute value i tried to change -cosx to cosx which ended up as

-7cosx

and then doing -7cos(3pi/2)-(-7cos(0))

but this was incorrect. Can anyone give me any hints?

thanks

 

[STEMucator]

Homework Statement

∫(0 to 3pi/2) -7|sinx|dx

Homework Equations

The Attempt at a Solution

I am not sure how to treat it as it has an absolute value

i assumed that you could remove the -7 to get

-7∫|sinx| dx

then integrate sinx into -cosx but since there is absolute value i tried to change -cosx to cosx which ended up as

-7cosx

and then doing -7cos(3pi/2)-(-7cos(0))

but this was incorrect. Can anyone give me any hints?

thanks

Usually for absolute values you have to break things into cases depending on what values of x you're integrating over.

Notice that from 0 to 3π/2, x≥0? If x≥0, then |sinx| = sinx.

Consider integrating from x=-2 to x=-1, x<0. If x<0, then |sinx| = -sinx.

 

[CAF123]

Homework Statement

∫(0 to 3pi/2) -7|sinx|dx

Homework Equations

The Attempt at a Solution

I am not sure how to treat it as it has an absolute value

i assumed that you could remove the -7 to get

-7∫|sinx| dx

then integrate sinx into -cosx but since there is absolute value i tried to change -cosx to cosx which ended up as

-7cosx

and then doing -7cos(3pi/2)-(-7cos(0))

but this was incorrect. Can anyone give me any hints?

thanks

Sketch the graph of y = sinx from 0 to 3pi/2. From this, can you see what g = |sinx| would look like? Do you then see why integrating sinx to simply -cosx is wrong?

 

[Mark44]

Usually for absolute values you have to break things into cases depending on what values of x you're integrating over.

Notice that from 0 to 3π/2, x≥0? If x≥0, then |sinx| = sinx.

This is NOT true. If 0 ≤ x ≤ π, then |sinx| = sinx, but for π ≤ x ≤ 2π, sin(x) ≤ 0.

Consider integrating from x=-2 to x=-1, x<0. If x<0, then |sinx| = -sinx.

This is not true, either. There are infinitely many intervals for which x < 0 but sin(x) ≥ 0.

 

[doctordiddy]

I tried doing it like this
http://answers.yahoo.com/question/index?qid=20080827185408AAwN43j

but i get -7cos(3pi/2)-7cos(0) which is still incorrect. Am i just making an algebra mistake or am i missing any extra steps (chain rule etc..)?

 

[CAF123]

It is still not right. Can you use |sinx| = sinx if sinx ≥ 0 and |sinx| = -sinx if sinx < 0? This is key to solving the problem.