How comes that ( after Desmos graphing calculator) the " area function" of f(x)=x² gives back negative values when x<0?

Question

I'm currently trying to understand basic calculus concepts.

I asked to Desmos graphing calculator to draw the graph

(1) of f(x) = x²

(2) of the " area function" or " accumulation function" of function f, that I defined as the integral from 0 to x of f(t)dt.

What astonishes me is that the curve of the area function is below zero for x<0, while the curve of the original function is always above the X-axis.

If my " area function " represents the area under the curve of f and the X axis ( limited by x=0 and the vertical line passsing through x), how comes that the values of the area function are negative for x < 0?

In order the " area function" to correspond to its informal concept , shouldn't we define it as

A(x)= absolute value of ( integral from 0 to x of f(t)dt ) ?

Answer 1

Integrals explicitly are signed areas. Your original function is always positive, and if you're integrating left-to-right, so $x_1>x_0$: $$\int_{x_0}^{x_1}f(x) dx >0$$ but if you're integrating right-to-left, so $x_1<x_0$: $$\int_{x_0}^{x_1}f(x) dx <0$$ The area under your original function between, say $x_0=-2$ and $x_1=0$, is $F(0)-F(-2)$, which you'll find is positive.

What you're plotting is a primitive function $F$, which has the useful property that the signed area under $f$ between $x_0$ and $x_1$ is $F(x_1)-F(x_0)$. Taking absolute values as you propose destroys this useful property.

Answer 2

It's normal. You have negative values for x<0 Intégration of f(x)=x^2 from a to b is always positif only if a < b

Answer 3

If $x<0$, the area is $\int_{x}^0 t^2\;dt$, which is opposite in sign from $\int_0^x t^2\;dt$.

The interval on the x-axis is $[x,0]$ when $x<0$.

Answer 4

Not really. It's kinda weird but still.

Area by integration is signed, the way displacement can be signed depending on which direction you decide positive to be. For example, if walking forward is a positive displacement vector, then walking backwards is negative displacement, even though for a layman walking and moonwalking $5m$ is the same thing.

Let's upgrade that thought. If displacement is signed then velocity should also be signed, due to $\dfrac{\text{negative displacement}}{\text{positive time}}$. This makes the reverse and drive options on a car different, since unaltered they take you in opposing directions.

So if you're doing integrals you know a displacement function is the integral of the corresponding velocity function. And therefore, if you allow your velocity function to have negative values(corresponding to a reverse drive), then the area under the curve with those conditions should also be signed. But that integral needs a bit of explanation.

Let's say I was recording the speed of a car on a highway. The distance covered by it, say from 1 to 2 seconds is positive(displacement also kinda by relative choice of which direction $+$ is). If you played the video backwards, and you decided to call the original displacement positive:

1) It will run from t=2 to t=1

2) The displacement you will see will be in the reverse direction, hence with an opposite sign.

Long story short, $\displaystyle\int_b^a f(x)dx = -\int_a^b f(x)dx$.

So your integral has the option of negatives if the upper bound is smaller than the lower bound. Check those intervals!