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4. Find the absolute maximum and minimum value of f(x)=S0g(t)dt on the interval [0, 5], where g(t) is given by the graph below. (Remark: The part of the graph over [3, 5] is a quarter circle, and the part of the graph over [0, 3] is a line going straight up and to the right.)
Note: I'm using "S" as the definite integral sign.

Meg,
Very nicely stated problem except the last in the Remark.  How can a straight line go straight up and to the right?
Ok, now I see it as this.  On the interval [0,3], the function is a straight line going from say [0,0] to [3,0] and on the interval [3,5] it is a circle with 2 units radius.  Does this sound right?
Correction from my previous comment.  I meant  [0,0] to [3,2] not  [0,0] to [3,0]
Okay sorry! What I meant by straight up and to the right is this:
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I hope that makes sense! & as for the intervals, the function is a straight line going from (0, -1) to (3, 2) on the interval [0, 3]. & yes, the other part sounds right! The interval [3, 5] looks like a circle with 2 units radius.

The first piece is just a straight line, y=x-1, for 0≤x<3.

If the second piece really is a quarter circle, you will need its equation. The equation for a circle centered at (3,0) of radius 2 is (x-3)² + y² = 4, so the quarter circle as a function is
y = sqrt(4-(x-3)²) = sqrt(-x²+6x-5), for 3≤x≤5.

So the piecewise function g(t) is
g(t) = { (t-1)                      0≤t<3
{ sqrt(-t²+6t-5)        3≤t≤5

Now f(x)=∫0x g(t) dt, which will also be piecewise continuous. It's the area between the graph of g(t) and the t-axis, counting areas below the t-axis as negative. To find the absolute extrema of f(x), you do not need to carry out the integral, assuming you know the area of a circle and of a triangle. For which x is the area smallest? For which x is the area largest? What are the min and max values of these areas?