We saw that the area O(f,a,b) of the region below the graph of a non-linear function f(x) can be approximated by summing the area of the best-fitting bars under the graph, hence, O=O1+O2+…On. The more bars we use (hence, the larger n), the better the approximation of the area. We now investigate the relation between this approximation and integrals.
Consider the previously used example with three strips.
The interval [a,b] is then split into three subintervals (of the same size) [a,x1], [x1,x2] and [x2,b]. The area of the first bar can be calculated as follows: O1=f(a)⋅(x1−a). In the same way we obtain the areas of the second O2=f(x1)⋅(x2−x1) and third bar O3=f(x2)⋅(b−x2). Conclusion:
O≈f(a)(x1−a)+f(x1)(x2−x1)+f(x2)(b−x2)(1).
Derivative: From Chapter 2 we know that for a derivative y′(x) of a function y(x) it holds that
y(x+Δx)−y(x)Δx≈y′(x).
When we replace in this expression the function y(x) by an antiderivative F(x) then we obtain
F(x+Δx)−F(x)Δx≈F′(x)=f(x),
which we can rewrite to
F(x+Δx)−F(x)≈f(x)Δx(2).
Relation to integral: Consider equation (2). If we take in this expression x=a and Δx=x1−a, such that x+Δx=x1, then
F(x1)−F(a)≈f(a)(x1−a)=O1.
Taking x=x1 and Δx=x2−x1, such that x+Δx=x2, gives
F(x2)−F(x1)≈f(x1)(x2−x1)=O2.
Taking x=x2 and Δx=b−x2, such that x+Δx=b, gives
F(b)−F(x2)≈f(x2)(b−x2)=O3.
Hence, we can rewrite equation (1) as
O≈O1+O2+O3≈(F(x1)−F(a))+(F(x2)−F(x1))+(F(b)−F(x2))=F(b)−F(a)(3).
Since F(b)−F(a)=∫baf(x)dx we have an approximation for the integral. Note that this approximation improves by increasing the number of bars. If the number of bars becomes infinite we can replace the ≈-signs by =-signs. This holds for both equation (2) and equation (3).
Conclusion: The area of the region enclosed by the graph of the function f(x), the x-axis and the lines x=a and x=b is equal to the integral ∫baf(x)dx.
Consider the previously used example with three strips.
The interval [a,b] is then split into three subintervals (of the same size) [a,x1], [x1,x2] and [x2,b]. The area of the first bar can be calculated as follows: O1=f(a)⋅(x1−a). In the same way we obtain the areas of the second O2=f(x1)⋅(x2−x1) and third bar O3=f(x2)⋅(b−x2). Conclusion:
O≈f(a)(x1−a)+f(x1)(x2−x1)+f(x2)(b−x2)(1).
Derivative: From Chapter 2 we know that for a derivative y′(x) of a function y(x) it holds that
y(x+Δx)−y(x)Δx≈y′(x).
When we replace in this expression the function y(x) by an antiderivative F(x) then we obtain
F(x+Δx)−F(x)Δx≈F′(x)=f(x),
which we can rewrite to
F(x+Δx)−F(x)≈f(x)Δx(2).
Relation to integral: Consider equation (2). If we take in this expression x=a and Δx=x1−a, such that x+Δx=x1, then
F(x1)−F(a)≈f(a)(x1−a)=O1.
Taking x=x1 and Δx=x2−x1, such that x+Δx=x2, gives
F(x2)−F(x1)≈f(x1)(x2−x1)=O2.
Taking x=x2 and Δx=b−x2, such that x+Δx=b, gives
F(b)−F(x2)≈f(x2)(b−x2)=O3.
Hence, we can rewrite equation (1) as
O≈O1+O2+O3≈(F(x1)−F(a))+(F(x2)−F(x1))+(F(b)−F(x2))=F(b)−F(a)(3).
Since F(b)−F(a)=∫baf(x)dx we have an approximation for the integral. Note that this approximation improves by increasing the number of bars. If the number of bars becomes infinite we can replace the ≈-signs by =-signs. This holds for both equation (2) and equation (3).
Conclusion: The area of the region enclosed by the graph of the function f(x), the x-axis and the lines x=a and x=b is equal to the integral ∫baf(x)dx.