Like

Report

Draw a graph of $ f $ and use it to make a rough sketch of the antiderivative that passes through the origin

$ f(x) = \dfrac{\sin x}{1 + x^2} $, $ \quad -2\pi \leqslant x \leqslant 2\pi $

See the explanation for solution

You must be signed in to discuss.

So here we're looking to draw the graph and make a rough sketch of the anti derivative as it passes through the origin. So given the sin of X over one plus X square, I'm negative two pi to pie. Or looking at is this region right here. So what we're gonna do is um recognize that our anti derivative, it's going to have zero slope. So it's going to be um like this, the area under the curve is very little and then at this point the area under the curve is negative. So that means that we're going to have a decreasing slope and then the area bullock curve is positive right here. So it's going to be increasing at that point. Um And then it's going to reach a maximum point at zero right here and then um at the zero point because that's the maximum this is going to come down and go toward zero. So what we're going to have is something like this where it comes up like a little hill and then it goes down. Um So an example, this would be like 1/1 class X squared. Something like this would be an example, however, it might dip a little bit lower. Um So it might be something like like this, but maybe we could expect one um be a little bit shifted up more alright, five. So it's something like this would be a little bit more accurate um in representing the anti derivative of this function.

California Baptist University