Question

Help please!(:
Graph the following intergrands and use the areas to evaluate the integrals.
-problems below.

Answer 1

\[1.) \int\limits_{-4}^{0}\sqrt{16-x^2} dx\]
\[2.) \int\limits_{-1}^{1} (1-|x|) dx\]

Answer 2

for the first one just use the substitution x=4sin(u)
the second one you can devide the integral into two intervals:
from (-1 to 0) \[\int\limits_{-1}^{0} 1-(-x)\]
and from (0 to 1) \[\int\limits_{0}^{1} 1-x\]

Answer 3

@pdd21 when the problem says "then use areas to evaluate the integrals", what is it exactly asking you to do?

Answer 4

IDK, that's all the question asks.. -.-

Answer 5

I think you meant find the area by evaluating the integral

Answer 6

It could be that the question wants you to realize what the area represents (i.e. what shape do you have)?

Answer 7

I think @SithsAndGiggles could be right.

Answer 8

That's what I'm thinking because it says "Graph the following intergrands..."

which suggests a geometric approach to figuring out the integrals.

Answer 9

\(\sqrt{16-x^2}\) graphed over the interval (-4,0) is a quarter-circle with radius 4.

\(1-|x|\) graphed over (-1,1) is a triangle with base 2 and height 1, I think. (trying to picture this mentally)

Answer 10

well said @SithsAndGiggles
I think these integrals can be easily found by evaluating the areas under the graphs

Answer 11

okay. I drew them out actually, how do I evaluate them? @Ahmad1

Answer 12

@pdd21, What are the areas of a quarter-circle and triangle?

Answer 13

\[A=(\frac{ 1 }{ 4 } ) \pi r^2\]
\[A=(\frac{ 1 }{ 2 }) b*h\]
@SithsAndGiggles

Answer 14

okay dude @pdd21 now just find the lengths from the graphs
e.g. the radius of the circle is 2.

Answer 15

lengths?

Answer 16

@Ahmad1

Answer 17

I am sorry I mean after you graphed them...
you can figure that the radius is 4 (not 2)
so the first integral is .5 pi r^2 = .5*16*pi = 8pi
the second one you have the base =2 and the height is 1 thus the area =.5*2*1=1

Answer 18

And that would be the answer? or would we still have to evaluate the integral? @Ahmad1

Answer 19

no my friend @pdd21 the definite integral for some function=the area between the graph and the x=axis, and I already found it
the first integral = 8pi and the second one =1

Answer 20

hmmm, I guess i over thought the problem I thought it wasn't as simple as how we did that.@Ahmad1