Integration (1 Viewer)

[Lukybear]

Find the exact area between the curve y=-e^-x, the x axis and the lines x=1 and x=4

From integration i found
A= 1/e^4 - 1/e

where it will yeild a negative answer.

My question is, if it asks to be expressed in "e" form, how will you now that it is infact 1/e-1/e^4.

 

[annabackwards]

You have to put the absolute value sign around the answer because it's below the curve.
So you're answer is |1/e-1/e^4| = +/-(1/e - 1/e^4)
You'd take the negative answer ie -(1/e - 1/e^4) = 1/e^4 -1/e so that it's a positive answer.
If you want to put a reason in brackets, just write (Area > 0)

 

[Lukybear]

I see, thanks.

 

[gurmies]

You have to put the absolute value sign around the answer because it's below the curve.
So you're answer is |1/e-1/e^4| = +/-(1/e - 1/e^4)
You'd take the negative answer ie -(1/e - 1/e^4) = 1/e^4 -1/e so that it's a positive answer.
If you want to put a reason in brackets, just write (Area > 0)

Similarly, you can just switch the limits =]

 

[annabackwards]

Similarly, you can just switch the limits =]

Oh, i never thought of that XD

 

[jet]

Similarly, you can just switch the limits =]

Ahh, but you can't really justify that in relation to an area like you can with absolute values.

In order to completely ensure that you're correct, one would need to graph the curve, and shade the area, to show the marker that you are completely aware of what you are actually doing.