Question

Integral calculus question

Answer 1

How would you approach this?

Answer 2

Well, for part a, remember that an integral can be visualized as the area under the curve. Since this "curve" is simply some straight lines, you can look at the graph and find out how the areas necessary by some simple geometric formulas.

Let's start with the easiest one. What's \(g(0)\)?

Answer 3

Wait actually, is it 0?

Answer 4

And g(1) = 2?

Answer 5

Not quite. You would be thinking of \(g(1)\) (which is 2). To figure out \(g(0)\), let's look at the definition of the function \(g(x)\). The definition is\[g(x)=\int_0^xf(t)\,dt\]When \(x=0\), we have\[g(0)=\int_0^0f(t)\,dt\]What's any integral from 0 to 0?

Answer 6

0

Answer 7

Perfect. So \(g(0)=0\), and as you already figured out, \(g(1)=2\). Now by finding the area under the graph between 0 and 2, what is \(g(2)\)?

Answer 8

Is it 5?

Answer 9

That's correct. Now \(g(3)=\,\,?\)

Answer 10

g(3) = g(2) + 2
g(3) = 7
Can i add them like this?

Answer 11

and g(6) = g(3) - 4?
so g(6) = 3?

Answer 12

That's exactly how you would add them, and both of those answers look correct.

Answer 13

Ok I get it now thanks.
but how do you find where g is increasing?

Answer 14

Intervals are the inverse functions of derivatives right? Does that mean something?

Answer 15

It's increasing whenever you're adding more area underneath the curve. So whenever your graph is above the horizontal axis, the function \(g(x)\) is increasing.

Answer 16

The interval is increasing from 0 to 3, and decreasing from 3 to 7 then

Answer 17

And the maximum value is where it changes from above the x-axis to below right?
So part c would be at x = 3?

Answer 18

Right again.

Answer 19

Ok thanks you were a big help :]