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.

# Functions defined by integrals: challenge problem

Solving for where a function defined by an integral equals 0. Created by Sal Khan.

## Want to join the conversation?

• Couldn't you also go from -7 to positive 5?
• Correct me if i'm wrong anyone but I believe we had to start from -5 as one of the conditions of the question. (Seen in top integral finding area from -5 to a value of x). Sorry if this wasn't explained very well
• -
How does Sal know it's a quarter circle? I mean, if it were a written test I'd probably have to prove both purple areas are equal, and that involves proving they have the same "format". Is there a way to spot a part of a circle just by seeing it?
(I'm considering the existence of infinite ways for going from (1;1) to (5;4), ranging from a straight line to a 90 degrees part of a circle).
• The thing I noticed was that the radius at t = -3 was 4 and the radius from (-3, 0) to (1, 0) was also 4, then the radius from (1, 0) to (5, 0) is ALSO 4 and the radius at x = 5 is 4. A curve that has a `constant radius` is following a segment of a circle. That was enough justification in my opinion for assuming a quarter circle for the purposes of solving these exercises--this curve was looking like a piece of a circle, and had a constant radius at the points I checked. If you really wanted to check the curve at another point, the circle passes through or very close to (-1, -3.5). Using the distance formula, this gives a radius ≈ 4.03, which is very close to 4, and if you squint at the curve, the 3.5 is actually an over-estimate, but at this scale, it is difficult to give more decimal places. (In order to be a radius of 4, the coordinate would be ≈ 3.464 --an approximation of √12 , and it isn't possible to derive that much from the drawing.) The other quarter circle passes through or very close to (3, 3.5). This point also gives a radius of ≈ 4.03. A third checkpoint is (1.5, 2), which is also ≈ 4.03. For the purposes of evaluating these integrals, that is close enough.
• Could I integrate from -5 to -7 i.e. going left on the x-axis?
• Yes, that would work. However, some instructors don't like that and would mark off for it. If you reverse the position of the bounds, the answer will be the negative of the original form. Thus,
∫ f(x) from a to b = − ∫ f(x) from b to a
• Negative area is related to whether or not the function is below the x-axis, right? So even if I have negative xs-values on my function, the area between the function and the x-axis would be positive if the function was positive on that "negative x interval", right?
(1 vote)
• The interpretation of an area below the x-axis depends on what is being represented by the function. For example, if the area represents something like profits and a negative area represents losses, then that would be subtracted. But, if the area represented, say, the area of a shape to cut a fabric, then the area below the x-axis would still be a physical area and be treated as a positive area. So, whether the area below the x-axis is treated as a positive or negative depends on the context and what the function/integral is representing.
• I am having a hard time understanding how "x" is not called "t" since he is explaining x values on the t axis. Couldnt the integral just be called from -5 to "t" of f(t)dt? Why call it x?
• Indeed he could have again used the variable `t`, he decided to change the name of the variable to avoid confusion. while inside the integral `t` is the integration variable, and since this is a definite integral, the integration variable will disappear once the integral is evaluated. To prevent confusion of a variable `t` still existing after the integral was evaluated, Sal decided to change the name of the variable to `x`, to represent the variable once the integral was evaluated.
(1 vote)
• How would you go about doing a problem like this algebraically without a graph?
• Why couldn't I have done the Integral between -5 to 1 + Integral between 1 to 6?
(1 vote)
• Because you're not finding f(t)'s zeroes or anything like that: you're simply finding the zeroes of F(x). In this case, you want the total area of f(t) between itself and the x-axis to be nullified. Those two areas would indeed zero each other, but that won't answer the question (especially when not given an equation for f(t)), if that makes any sense. You need to find the values for F(x)'s zeroes without an equation, as Sal does in the video.
• what is integral from sin(-pi) from sin(pi)?
(1 vote)
• It would be zero also, via symmetry of the sin function. The negative area between -pi and 0 is equal to the positive area between 0 and pi.
See https://www.desmos.com/calculator/hu5yauacj3