Definite Integrals

For a function f (x) that is continuous in the closed interval [a, b], it is possible to calculate the integral between the limits, a and b. An integral calculated between two limits is called a definite integral.

Definite integrals – notation

A definite integral usually produces a value, unlike an indefinite integral, which produces a function.

Definite integrals are depicted in the same way that indefinite integrals are depicted with the addition that the limits are added as subscript and superscript on the Integration sign. For example, if we want to integrate, $x^2$ between the limits 5 and 8, the corresponding Notation would be

${\int }_5^8{x}^{2}dx$

Solving definite integrals

How do you solve definite integrals? To solve for definite integrals, follow the following procedure:

1) Write down the definite integral with its limits in the form,${\int }_a^{b}f’\left(x\right)dx$

2) Integrate the function f '(x) the same way you would for an indefinite integral to find out f (x). Do not include the Integration constant, C. Write the result in the form,${\left[f\right(x\left)\right]}_a^b$

3) Now evaluate f (x) between the given limits: f (b) - f (a) This gives you the final value.

Example 1

Evaluate ${\int }_1^{7}5x^{2}dx$

Solution 1

${\int }_1^{7}5x^{2}dx$

= ${[5x^3/3]}_1^7$

= (5/3 × 7³) - (5/3 × 1³)

= 570

Finding the area under a curve

Integration is a very useful tool for Finding the Area under a graph. In the above example, we are Finding the Area enclosed between the x-axis and the curve f (x) = 5x² between x = 1 and x = 7. We can represent the above example graphically.

Figure 1. Graph depicting f (x) = 5x², to find the area enclosed between x = 1 and x = 7, Nilabhro Datta - StudySmarter Originals

The curve in the above graph represents f (x) = 5x². As mentioned, the value of the definite integral between 1 and 7 gives the area enclosed between the curve and the x-axis between x = 1 and x = 7.

Example 2

Evaluate ${\int }_{0.5}^1\cos \left(x\right)dx$Note - x is in Radians

Solution 2

${\int }_{0.5}^{1}c\mathrm{os}\left(x\right)dx$

= ${[\sin (x)]}_{0.5}^1$

= sin (1) - sin (0.5)

= 0.841-0.479

= 0.362

Like the previous example, the above value gives us the area enclosed between the curve y = cos (x) and the x-axis between x = 0.5 and x = 1. Check the following image for a clear demonstration.

Figure 2. Graph depicting f (x) = cos (x), to find the area enclosed between x = 0.5 and x = 1, Nilabhro Datta - StudySmarter Originals

Example 3

Given ${\int }_1^5$ (2Px + 7) dx = 4P², show that there are two possible values of P. Find these values.

Solution 3

${\int }_1^5$ (2Px + 7) dx

${[Px²+7x]}_1^5$ = (25P + 35) - (P + 7)

= 24P + 28

Now,

24P + 28 = 4P²

=> 6P + 7 = P²

=> $P^2-6P-7=0$

=> (P + 1) (P - 7) = 0

Therefore, the value of P can be either -1 or 7.

Example 4

Find the closed area bounded by the curve y = x (x - 5) and the x-axis.

Solution 4

To find the area bounded by the curve and the x-axis, let us find the points at which the curve intersects with the axis, ie. where y = f (x) = 0

f (x) = x (x - 5) = 0

=> x = 0 or x = 5.

So the curve intersects the x-axis at (0, 0) and (5, 0).

0 and 5 serve as the Lower and Upper Bounds for our definite integral.

So, total area = ${\int }_0^5$ (x) (x - 5) dx

$$\begin{gathered} ={\int }_0^5(x²-5x)dx \\ ={[x^3/3-5x²/2]}_0^5=(125/3-125/2)-(0-0) \\ =-20.83 \end{gathered}$$

A negative area bounded by a curve

In the above example, we have the area coming out to be a negative value. What does this mean?

It implies that the area enclosed by the curve and the x-axis falls below the x-axis, i.e. the negative side of the x-axis.

If we draw the curve y = f (x) = x (x - 5), we obtain the following curve.

Figure 3. Graph depicting f (x) = x (x - 5), to find the area enclosed between the curve and the x-axis, Nilabhro Datta - StudySmarter Originals

As we can see here, the area bounded by the curve falls below the x-axis.

What if we want to find the entire magnitude of the area enclosed between a curve and the x-axis when some of it falls above the x-axis, and some of it falls below the x-axis? In these cases, we would need to find both the areas individually and add their magnitudes together without taking into account their sign.