The Definite Integral, how it relates to the area under the curve and summations.

Hello fello mathematicians,

I have been learning calculus recently for the first time, independently, and I have hit a wall. In studying the definite integral in depth, I have failed to see how it calculates the area under a curve. I understand that the area under a given curve bounded by two points is the sum of all rectangles as their width approaches zero. I can see that as a sum which is equal to the area under a curve. But I dont see how takes the difference between the anti-derivative at 2 points is equivalent to the area under the curve. I have read two different textbooks and how they approach it, and even online articles, and I understand everything but this relation. May someone please help or provide a better proof or intuitive explanation. The difference is two anti-derivatives of a function to me seems like just taking the difference of two infinitesimally small areas(rectangles). Wouldnt we have to add up all those infinitesimally small rectangles bounded by two points to find the area? I know that is true, but how is that the same as taking the difference between two anti-derivatives at points b and a?

Re: The Definite Integral, how it relates to the area under the curve and summations.

Quote:

Originally Posted by

**adamreddy** Hello fello mathematicians,

I have been learning calculus recently for the first time, independently, and I have hit a wall. In studying the definite integral in depth, I have failed to see how it calculates the area under a curve. I understand that the area under a given curve bounded by two points is the sum of all rectangles as their width approaches zero. I can see that as a sum which is equal to the area under a curve. But I dont see how takes the difference between the anti-derivative at 2 points is equivalent to the area under the curve. I have read two different textbooks and how they approach it, and even online articles, and I understand everything but this relation. May someone please help or provide a better proof or intuitive explanation. The difference is two anti-derivatives of a function to me seems like just taking the difference of two infinitesimally small areas(rectangles). Wouldnt we have to add up all those infinitesimally small rectangles bounded by two points to find the area? I know that is true, but how is that the same as taking the difference between two anti-derivatives at points b and a?

In order to prove that the area under a curve is equal to the definite integral, one first needs to know the Mean Value Theorem. The Mean Value Theorem states that for any curve, if you choose two points and draw a chord between them, the gradient of this chord will be equal to the gradient of a tangent to the curve somewhere in between those points. In other words: $\displaystyle \displaystyle \frac{f(\beta) - f(\alpha)}{\beta - \alpha} = f'(\gamma)$ for some $\displaystyle \displaystyle \gamma \in [\alpha, \beta]$. If we rearrange this, we get $\displaystyle \displaystyle (\beta - \alpha)f'(\gamma) = f(\beta) - f(\alpha) $, this will be important for later.

Now, start by splitting the region you are evaluating the integral over into $\displaystyle \displaystyle n$ sub intervals, and also mark in their midpoints, such that $\displaystyle \displaystyle a = x_0 < m_1 < x_1 < m_2 < x_2 < \dots < m_n < x_n = b$ and approximate the integral using rectangles of the same length as one of these subintervals, i.e. $\displaystyle \displaystyle x_i - x_{i - 1}, i = 1, 2, 3, \dots , n$ and width evaluated at the midpoint of each subinterval, i.e. $\displaystyle \displaystyle f(m_i)$. Then the area of each rectangle is $\displaystyle \displaystyle (x_i - x_{i - 1})f(m_i) $ and the total area can be approximated by $\displaystyle \displaystyle \sum_{i = 1}^n{(x_i - x_{i - 1})f(m_i)} $.

In order to improve on this approximation, we need to increase the number of subintervals, to make the length of each subinterval smaller and to decrease the error. So the exact area is the limit of this sum as we make $\displaystyle \displaystyle n \to \infty $.

Therefore $\displaystyle \displaystyle A = \lim_{n \to \infty}\sum_{i = 1}^{n}{(x_i - x_{i-1})f(m_i)} $, and since we are making the subintervals extremely small, the midpoint ends up being the ONLY point between the two endpoints of each subinterval, which means we can simplify using the rearranged form of the Mean Value Theorem that is given above. So

$\displaystyle \displaystyle \begin{align*} A &= \lim_{n \to \infty}\sum_{i = 1}^{n}{(x_1 - x_{i - 1})f(m_1)} \\ &= \lim_{n \to \infty}\sum_{i = 1}^n\left[F(x_i) - F(x_{i - 1})\right] \textrm{ where }F(x)\textrm{ is an antiderivative of } f(x) \\ &= \lim_{n \to \infty}\left\{ \left[F(x_1) - F(x_0)\right] + \left[F(x_2) - F(x_1)\right] + \left[F(x_3) - F(x_2)\right] + \dots + \left[F(x_n) - F(x_{n-1})\right] \right\} \\ &= \lim_{n \to \infty}\left[F(x_n) - F(x_0)\right] \\ &= \lim_{n \to \infty}\left[F(b) - F(a)\right] \\ &= F(b) - F(a) \end{align*} $

Hope that helped :)

Re: The Definite Integral, how it relates to the area under the curve and summations.

Also, if intuition is a priority, it can help to consider A(t), the cumulative area function for a given function f(x).

E.g., A(b) gives the area under f(x) from as far 'left' as f(x) is above the x axis, and up as far as the value b on the x axis.

Just by considering the nature of a cumulative function you can see that the area under f(x) from a up to b along the x axis must be A(b) - A(a).

So if we can find A(t) from f(x), we're done.

(Hope I'm not being pedantic calling it A(t). You can probably safely say A(x) if preferred.)

The trick is, the derivative of A(t) turns out to be f(x). See Fundamental theorem of calculus - Wikipedia, the free encyclopedia for an intuitive grasp of why the rate of increase in the area (as we travel rightward along the x axis) is always the height of the curve.

The trouble, then, is that the derivative function is many-one, so there are any number of cumulative area functions that will differentiate to give f(x). So the best we can say for A(t) is: the anti-derivative of f(x) plus an unknown constant C. But then another trick: C is the same for A(a) as for A(b), so they cancel!

Re: The Definite Integral, how it relates to the area under the curve and summations.

that was a good proof. im surprised your the first one to mention it too me. That is probably the most sound geometric proof I have seen. Riemann sums is easy to understand but I think to fully take grasp of its relation to the Definate integral you must incorporate the mean value theorem as well.