Explain me definite integral...

Question

So, as the derivative is a tangent of an angle between the $x$-axis and the corresponding tangent line, how can we represent an indefinite integral, and why is the area (definite integral) of, for example, $y=x^2+1$ from $-1$ to $1$ observed below the parabola?

Answer 1

The same way the derivative of a function is defined as $$\lim_{\delta x\to 0} \frac{f(x+\delta x) - f(x)}{\delta x}$$

The definite integral $\displaystyle\int_a^b f(x) \ dx$ is defined as $$\lim_{n\to\infty} \sum_{i=1}^n f(x_i) \delta x$$

where $\delta x = \dfrac{b-a}{n}$ and $x_i = a + i\delta x$.

The above sum is quite intuitive if you think about it. Suppose you have a curve. The area under that curve can be approximated by drawing $n$ very small rectangles of equal width up to the point where the touch the curve. If you want to find the area under the curve over $[a,b]$ then it can be approximated as the sum of all $n$ little rectangles which you've drawn under the curve. The width of the rectangle let us call $\delta x$ where $\delta x = \dfrac{b-a}{n}$ as it is the width of the interval divided by the number of divisions you want to approximate the area. The height of the rectangle will be $f(x)$ evaluated at that point i.e. $f(a+i\delta x)$ where $i\in\{1,2,\cdots ,n\}$. As we let $n\to\infty$ we get a better and better approximation for the area under the curve.

If this explanation isn't very clear/ you want to know more: google Riemann sum.

EDIT: EXAMPLE

Using the above definition for the definite integral, calculate $\int_0^2 x^2 \ dx$.

We have our lower limit, $a$, is 0 and our upper limit, $b$, is 2. Now, we want to approximate the area under the graph of $x^2$ over $[0,2]$ so we split up the area under the graph into $n$ rectangles of equal width $\dfrac{b-a}{n} = \dfrac{2-0}{n} = \dfrac{2}{n}$.

Therefore, our approximation for the area under the graph is $$\int_0^2 x^2 \ dx \approx \sum_{i=1}^n f(x_i) \delta x = \sum_{i=1}^n (0 + i\delta x)^2 \frac{2}{n} = \sum_{i=1}^n \left(i \frac{2}{n}\right)^2 \cdot \frac{2}{n}$$

Computing the sum above on the RHS using standard results:

$$\sum_{i=1}^n \left(i \frac{2}{n}\right)^2 \frac{2}{n} = \frac{8}{n^3} \sum_{i=1}^n i^2 = \frac{8}{n^3} \frac{n(n+1)(2n+1)}{6}$$

Simplify the term on the RHS:

$$\frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{8}{n^3} \cdot \frac{2n^3 + 3n^2 + n}{6} = \frac{8}{6} \cdot \underbrace{2 + 3n^{-1} + n^{-2}}_{\to 2}$$

as $n\to\infty$. So our final result is $\frac{8}{3}$ (which agrees with what we would get from using integration).

Answer 2

The area enclosed between those and $x$-axis