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# Finding derivative with fundamental theorem of calculus: x is on both bounds

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)
How do you apply the fundamental theorem of calculus when both integral bounds are a function of x. Created by Sal Khan.

## Want to join the conversation?

• can someone explain why he added the times 2x at the back of F'(x)? I can't seem to understand that. I did however understand why he had to in the previous video. but in this case. i don't see the need.
• Because our upper bound was x², we have to use the chain rule to complete our conversion of the original derivative to match the upper bound. The derivative of x² is 2x, and the chain rule says we need to multiply that factor by the rest of the derivative.
• What would happen if you have intervals of numbers rather than x and x^2? For example if you have over the interval of 0 to 1 of the function cos(t)/t.
• Then, F(x) would be a constant since the input x is not used in the expression. If you take the derivative of a constant, the result is 0, meaning F'(x) would be 0.

I hope this helps!
• I don't really get how c is constant if its value depends on x. If by constant he means for a particular value of x, then wouldn't the bounds of the original integral also be constant and so wouldn't there be no point in splitting the integral into two parts in the first place?
• This is one of those confusing contraptions where the underlying function -- the one portrayed in the graph -- is a function of t, not a function of x. The constant c is a t-value, not an x-value. It doesn't depend on x. If you aren't getting how F(x) can be a function of x when we're doing an integral of a function of t, it may help to go back over the earlier videos in this section on the fundamental theorem.
• At , why does Sal take the derivative of x^2? Didn't we already apply the Fundamental Theorem of Calculus, or does the fundamental theorem also state that we must take the derivative of the upper-bound?
• What about when the lower and upper limits of the integral both contain a variable for instance the integral from 3x to x^2 of 1/(2+e^t) ? How would you solve that problem?
• You simply do the integral in the normal way, and then substitute in the limits which are functions of x. You end up with an expression which is a function of x. This is quite reasonable, if you think about it -- a definite integral gives you the area below the curve between the two specified limits. If the limits depend on x, then the area is not going to be constant, but will also depend on x.

integral_(3x)^(x^2) 1/(2+e^t) dt
= [-ln(2+e^t) / 2 + t/2]_(3x)^(x^2)
= -ln(2+e^(x^2)) / 2 + (x^2)/2 + ln(2+e^(3x)) / 2 - 3x/2.
• How to calculate the indefinite integral ∫cos(x)/x dx?
• Instead of using c, would you be able to take the integral from 0 to x^2 - x ?
• I just solved it using 0 instead of c and it gave me the same answer, so I think it's acceptable.
• At Sal cancelled one x out of x^2 in the denominator with the only x in the numerator and wrote simply x but why can't x be negative?So i think it's reasonable to write ABSOLUTE VALUE OF X instead of simply x.Please help!
(1 vote)
• Remember x is a variable, a placeholder for any value.
Now . . . .
Simplify x²/x. The answer is x. In no way have we limited x to be non negative.
In fact, by placing absolute values around the x, as you suggest, you are actually saying, "I don't care if the value x takes on is negative, make the result positive."
Example: let x = -5, then x²/x = (-5)²/-5 = (-5)(-5)/-5 = -5.
If we did it your way we would be saying that (-5)²/-5 = 5, and that is not correct.