/sci/ - Science & Math

Why does the integral operation give the area under the curve of a function? For example, [math]f(x)=2x[/math] over the domain a to b is equal to [math]F(x)=x^2[/math] evaluated at [math]F(b) − F(a)[/math]. Why does the operation [math]\frac{x^{n+1}}{n+1}[/math] give the area under the curve of a given function? This is not intuitive at all. If I take the sum of all the infinitesimally small rectangles underneath the curve of a function over the domain a to b, that is the same as performing the operation [math]\frac{x^{n+1}}{n+1}[/math] on the function and evaluating the function at a and b and subtracting the former from the latter. Not one of my calculus professors have been able to explain this phenomenon without muh fundamental theorem of calculus but all that does is relate the derivative to the integral by saying the integral is defined as the antiderivative. It doesn't explain why it gives the area under the curve of a given function.

Why does the integral operation give the area under the curve of a function? For example, [math]f(x) = 2x over the domain a to b is equal to F(x) = x^2 evaluated at F(b) - F(a)[/math]. Why does the operation [math]\frac{x^{n+1}}{n+1}[/math] give the area under the curve of a given function? This is not intuitive at all. If I take the sum of all the infinitesimally small rectangles underneath the curve of a function over the domain a to b, that is the same as performing the operation [math]\frac{x^{n+1}}{n+1}[/math] on the function. Not one of my calculus professors have been able to explain this phenomenon without muh fundamental theorem of calculus but all that does is relate the derivative to the integral by saying the integral is defined as the antiderivative. It doesn't explain why it gives the area under the curve of a given function.