Applications of Integration

Topic Content:

Areas Under Curves:

Consider the area A of the figure bounded by the curve y = f(x), the x-axis and the y vertical lines through x = a and x = b and (where b > a)

If x = b then \(\scriptsize A_b = \int\limits_{x=b} y dx \)(the value of the integral and hence the area up to b )

and if x = a then \(\scriptsize A_a = \int\limits_{x=a} y dx \)(the value of the integral and hence the area up to a ).

Since b > a, the difference in these two areas \(\scriptsize ( A_b \: – \: A_a) \)gives the required area A.

A = \(\scriptsize \int\limits_{x=b} y dx\; – \; \int\limits_{x=a} y dx \)

A = \(\scriptsize \int_{a}^{b} y \: dx \)

This is called the definite integral of f(x) with respect to x, between the limits a (the lower limit) and b (the upper limit). It is a function of a and b .

The arbitrary constants disappear in the subtraction.

Example 5.8.1: 

Find the area bounded by the curve y = 3x² + 6x + 8, the x-axis and ordinate x = 1 and x = 3.

Solution: 

A = \( \scriptsize \int_{1}^{3} y dx = \int_{1}^{3} \left ( 3x^2 + 6x + 8 \right) dx \)

A = \( \left[ \frac{3x^3}{3} + \frac{6x^2}{2} + \scriptsize 8x \right]_1^3\)

= \( \left[\scriptsize x^3 + 3x^2 + 8x \right]_1^3\)

A = \( \left[\scriptsize (3)^3 + 3(3)^2 + 8(3) \right] \; – \; \left[\scriptsize (1)^3 + 3(1)^2 + 8(1) \right]\)

A = 78 – 12

A = 66 units²

Exercise: