Definite Integrals

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.

The area enclosed by the curve and the x-axis falling above the x-axis gives a positive value for ∫f (x) dx, and the area enclosed by the curve and the x-axis falling below the x-axis gives a negative value for ∫ f (x) dx.

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.