Question

Help...

Answer 1

Here's what i did so far,

I pulled the 3 out and then i tried to simplify the inside
so I got like (4+x)(4-x) and then i got stuck

Answer 2

Ermm, that shouldn't be necessary. You can't pull the 3 out because it is a term not a factor. You can integrate it separately as
\[\int\limits 3 +\sqrt{16-x^2} dx = \int\limits 3 dx + \int\limits \sqrt{16-x^2} dx\]

Answer 3

If you express
\[\sqrt{16-x^2} \rightarrow (16-x^2)^{1/2}\]
You can take antiderivative by thinking power rule and chain rule in reverse.

Answer 4

it says in terms of areas, so you aren't suppod to integrate it, you are supposed to recognize the shape of the graph

Answer 5

By the way, the shape of the area you are finding looks like this:

Answer 6

supposed*

Answer 7

right :)

Answer 8

@TuringTest ha, good point. That's even better! Geometry to the rescue!

Answer 9

so do you see what we're getting at here?

Answer 10

so i just have to take the antiderivative

Answer 11

no, the whole point is that you probably don't know how to integrate the second term at this stage in calc, it requires a trig sub

Answer 12

I like @TuringTest 's idea better. It's a quarter-circle on top of a rectangle. Ignore the calculus.

Answer 13

I just went by the way the problem was stated

Answer 14

Might be fun to do it both ways to verify . . .

Answer 15

look at the first integral\[\int_{-4}^03dx\]what will this area look like @Unam ?

Answer 16

hmm

Answer 17

draw the graph of y=3