Question

CALCULUS I | "Evaluate the following integrals by interpreting it in terms of areas"

Answer 1

\[a)~\int_{-1}^{4}(2x-1)dx\]\[b)~\int_{0}^{4}(2-\sqrt{16-x^2}dx\]

Answer 2

What exactly does "interpreting it in terms of areas" mean? Should I use left point, right point or midpoint intervals?

Answer 3

Speaking of intervals... what should I set \(n\) as?

Answer 4

I think your teacher wants you to break up the integrals into two pieces, then subtract the two areas

eg: this basic form
\[\Large \int_{a}^{b}(f(x)-g(x))dx = \int_{a}^{b}f(x)dx-\int_{a}^{b}g(x)dx\]

Answer 5

Ah. I see o:

Answer 6

So \(g(x)=1\) in this case?

Answer 7

yes

Answer 8

Okay... I am assuming I should get this when breaking up then.\[\int_{-1}^{4}(2x-1)dx=\int_{-1}^{4}(2x)-\int_{-1}^{4}(1)\]

Answer 9

I am not exactly sure how to handle the broken-up parts though... should I use \(\Sigma\) summation?

Answer 10

have you learned about antiderivatives?

Answer 11

Yes
Oops, I forgot the \(dx\) at the end it seems

Answer 12

Solving an integral \(\rightarrow\) antiderivative of integrated function?

Answer 13

I would use that method over Riemann Sums

Answer 14

if dy/dx = 2x then y = ???

Answer 15

\(y=x^2\) when \(\frac{dy}{dx}=2x\) because \((x^2)'=2x\)

Answer 16

yes, good

Answer 17

well +C but that +C will go away after you evaluate the endpoints

Answer 18

So then...\[\int_{-1}^{4}(2x)dx\rightarrow x^2\]and then...\[\int_{-1}^{4}(1)dx\rightarrow x\]

Answer 19

I am not sure if I should add a C to those or not...

Answer 20

no need because the C's will cancel

Answer 21

once you get to x^2, evaluate it at the endpoints and subtract

same for the other piece

Answer 22

Evaluate at... endpoints?

Answer 23

OH.
Well, I feel ridiculously stupid now.