Thread: A simple definite intergration problem

1. A simple definite intergration problem

I know it would sound stupid, but I am really confused.
Find the area under y=sinx from -π/2 to π/2. If I plug this into TI-83 calculator, I get 2. Which makes sense to me.

But if I do it by myself, the definite integral ∫sinx dx from -π/2 to π/2. I got zero on it, because the antiderivative of sinx is -cosx, so both the values for π/2 and -π/2 is zero, isn't it.

So this two way of doing this don't match. What's the problem?
Thank you.

2. Hi

This is not a stupid question

It depends if you are talking about the geometric area (always positive) or the algebraic area (negative when f(x) is negative)

The integration of sin x is 0 as you have calculated it

If you want to calculate the geometric area (which is 2), you should integrate the absolute value of sin x

3. Originally Posted by running-gag
Hi

This is not a stupid question

It depends if you are talking about the geometric area (always positive) or the algebraic area (negative when f(x) is negative)

The integration of sin x is 0 as you have calculated it

If you want to calculate the geometric area (which is 2), you should integrate the absolute value of sin x
Yes. I understand you. But I do the geometric area without a calculator, I still get 0. Antiderivative of sinx is -cosx, then -cosx from -π/2 to π/2 is -cos(π/2)-(-cox(-π/2))=0. where did I do wrong? I can't get 2 without a calculator.