How do I determine the area of a region bounded by a function and the x-axis if given y=cosx and interval is [0,π/2]? This has to do with the Fundamental Theorem of Calculus. Thanks in advance :)

Question

amistre64 · Answer

you integrate the function from 0 to pi/2

amistre64 · Answer

$$\int_{0}^{pi/2}f(x)dx=\left.F(x)\right|_{0}^{pi/2}$$
$$F(pi/2)-F(0)=Area.under.curve$$

amistre64 · Answer

do you know where cos(x) is derived from?

amistre64 · Answer

F(x) = ?

F'(x) = f(x) = cos(x)

anonymous · Answer

I don't know what you mean. cos(x) is given. I tried following the steps on the attached file but didn't understand how to get the "sin(x) on the second step. (please refer to attachment) Thanks.

amistre64 · Answer

there are 2 terms that are used for this, integration is one of them; and anti-differentation is the other.  Which one sounds more familiar to you?

anonymous · Answer

Well they are the same thing, right?

amistre64 · Answer

yes, but since your befuddled about what ive said i have to make sure i relay it to you in terms that you are capable of associating with

amistre64 · Answer

y = cos(x) might as well be read as:

y' = cos(x), find the y that got dervied

amistre64 · Answer

derived

amistre64 · Answer

in other words, underive cos(x)

anonymous · Answer

so y=cosx is not f(x)? It is f'(x)?

amistre64 · Answer

that is the best way to view it yes.  Call it by a name that you can see it for what it truly is.

amistre64 · Answer

f'(x) = cos(x)  is what you want to undo to get back to f(x)

anonymous · Answer

but then wouldn't f(x) be -sinx, not sinx, because the derivative is y' cos(x)=-sin(x)?

amistre64 · Answer

youre going the wrong way ..... let me try to draw it out :

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