If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 2011 Calculus AB free response #4b

Absolute maximum over an interval. Critical points and differentiability. Created by Sal Khan.

## Want to join the conversation?

• why is the derivative f(x) and not f(t)
• The Fundamental Theorem of Calculus states that if you are trying to find the derivative of g(x)= definite integral from a to x of f(t)dt, then you simply replace the x for every t in the inner function. Therefore, g'(x) = f(x).
• Is it possible to have a function that produce a circle and straight line?!
• Not a circle, because it would fail the verticle line test, but anything less than or equal to a semicircle is allowed, so are straight lines. I can't think of one single way to define this function, but it could be defined piecewise.
• No, this questions does not allow calculators (someone confirm that I am right).
• Discuss the role of constant of integration in indefinite integrals, why we don't involve it in definite integrals? (Be precise)
• Maybe I can try it in a more general example. Let f(x)=d/dx F(x
The indefinite integral ∫ f(x) dx = F(x) + c
Now, if I want to find the definite integral (area under the curve) ∫ f(x) dx from a to b, the fundamental theorem tells us that this is the same thing as the indefinite integral evaluated at the endpoints. So we have ( F(b) + c ) - ( F(a) + c ). The constant of integration is there in both minuend and the subtrahend. Now, what happens if we distribute the subtraction?
F(b) + c - F(a) - c. We can simplify this! The two constants are the same, so they subtract to 0. Thus, our definite integral simplifies to F(b) - F(a). Notice: no constant. Because of the canceling out, we tend to ignore the constant of integration when taking definite integrals because it will always simply subtract out.
• the problem asks for the absolute maximum and the |-8 - 2pi| is clearly the largest so should the value be x = -4 and not x = 5/2
• No. Absolute maximum means the highest point on the graph of g(x), not the point where the absolute value of g(x) is the highest.
• I think Sal's answer is right, but I'm not sure about the evaluation at x = -4... I think 2 pi should have been added, not subtracted.
The formula for g(x) is 2x + area under f(x) from 0 to -4. There is clearly more positive area than negative area, so it should be 2x + some positive number (2 pi).
I'm thinking that since Sal didn't directly calculate the area, he forgot to check signs? Am I wrong?
• It is subtracted because they reversed the normal way of expressing the interval which reversed the sign. (Integral from 0 to -4 instead of Integral from - 4 to 0) He explains this in the previous video.
• Why is the answer x=5/2, when the max on the graph is x=0?
• at x=0, f(x) has a maximum, but we are looking for the maximum of g(x)
• what should i do to solve a problem for this video that i dont know with out watching the video agian or getting the problem wrong does any one have any tips for me
• Post the problem and your solution and we can tell you whether you're correct.
• Hey Sal, I'm curious to know why you didn't g(3/2), where the straight line on the right side of this graph has a zero, would that be the absolute maximum? It seems to me that g(x) would be greater at that point than at g(5/2) because g(5/2) has negative area from integration lowering the number.
• At g(3/2) = 2* (3/2) + integral(f(t),t,0,3/2) = 6/2 + 9/4 = 21/4 or 5.25
at Sal shows that g(5/2) = 6, which is larger.

It makes since if you remember that a function will increase as long as its derivative is positive and the function decreases when the derivative is negative. To find the points where a function changes from increasing to decreasing or decreasing to increasing, can generally be found by setting the first derivative to 0. For this function that only happened at x = 5/2