Evaluating a Definite Integral

You might have noticed that an indefinite integral just has the integral symbol, $\int ,$ and after evaluating it you still have variables and an integration constant, for example$$\int x^2\mathrm{d}x=\frac{1}{3}{x}^3+C.$$Definite integrals, instead, have integration limits, like ${\int }_1^3,$ the end result is a number, and there are no integration constants, like

$${\int }_1^3{x}^2\mathrm{d}x=\frac{26}{3}$$

Evaluate

$${\int }_0^5{x}^2\mathrm{d}x$$

using the definition of definite integral.

Solution:

In this case, the function is $f(x)=x^2$ and the limits of integration are $a=0$ and $b=5.$ Knowing this, you can find $\mathrm{\Delta }x$ :

$$\begin{array}{rl}\mathrm{\Delta }x& =\frac{b-a}{N}\\ & =\frac{5-0}{N}\\ & =\frac{5}{N}.\end{array}$$

With this, you can use the definition of definite integral, so

$${\int }_0^5{x}^2\mathrm{d}x=\underset{N\to \mathrm{\infty }}{lim}\left[\sum _{i=1}^N{\left(x_i^{\ast }\right)}^2\left(\frac{5}{N}\right)\right].$$

You can use any point within each subinterval to evaluate the Riemann sum. For illustrative purposes, a right-endpoint approximation will be used. This will let you write

$$\begin{array}{rl}{x}_i^{\ast }& =a+i\mathrm{\Delta }x\\ & =i\frac{5}{N},\end{array}$$

which can be substituted back into the definite integral

$${\int }_0^5{x}^2\mathrm{d}x=\underset{N\to \mathrm{\infty }}{lim}\left[\sum _{i=1}^N{\left(i\frac{5}{N}\right)}^2\left(\frac{5}{N}\right)\right].$$

To evaluate the above limit begin by rewriting the expression to simplify it a little:

$$\begin{array}{rl}{\int }_0^5{x}^2\mathrm{d}x& =\underset{N\to \mathrm{\infty }}{lim}\left[\sum _{i=1}^N{\left(i\frac{5}{N}\right)}^2\left(\frac{5}{N}\right)\right]\\ & =\underset{N\to \mathrm{\infty }}{lim}\left[{\left(\frac{5}{N}\right)}^3\sum _{i=1}^N{i}^2\right].\end{array}$$

You can use the formula

$$\sum _{i=1}^N{i}^2=\frac{N(N+1)(2N+1)}{6}$$

to evaluate the sum, so

$${\int }_0^5{x}^2\mathrm{d}x=\underset{N\to \mathrm{\infty }}{lim}\left[{\left(\frac{5}{N}\right)}^3\left(\frac{N(N+1)(2N+1)}{6}\right)\right].$$

The above expression can be rewritten by expanding the product, that is

$${\int }_0^5{x}^2\mathrm{d}x=\underset{N\to \mathrm{\infty }}{lim}(\frac{125}{3}+\frac{125}{2N}+\frac{125}{6N^2}).$$

Finally, evaluate the above limit to find

$${\int }_0^5{x}^2\mathrm{d}x=\frac{125}{3}.$$

Evaluate

$${\int }_0^5{x}^2\mathrm{d}x$$

using the Fundamental Theorem of Calculus.

Answer:

Begin by finding the antiderivative of $x^2.$ This can be done using the Power Rule, so

$$\int x^2\mathrm{d}x=\frac{1}{3}{x}^3+C.$$

Next, you need to evaluate this antiderivative at both integration limits and subtract them. No matter what value of $C$ you choose, it will cancel out when doing the subtraction, so there is no need to include it when using the Fundamental Theorem of Calculus. That means the integral is

$$\begin{array}{rl}{\int }_0^5{x}^2\mathrm{d}x& =(\frac{1}{3}(5{)}^3)-(\frac{1}{3}(0{)}^3)\\ & =\frac{125}{3}.\end{array}$$

Evaluate the definite integral

$${\int }_0^{4}2x\mathrm{d}x.$$

Answer:

In this case you are trying to find the area below the linear function $f(x)=2x.$ Begin by taking a look at its graph.

Fig. 3. Graph of the linear function.

Note that the area below the function is a triangle with base 4 and height 8.

Fig. 4. The area below the function forms a right triangle.

Therefore, you can use the formula for the area of a triangle to find this area, so

$$\begin{array}{rl}A& =\frac{bh}{2}\\ & =\frac{4(8)}{2}\\ & =16.\end{array}$$

This means that the value of the definite integral is 16.

$${\int }_0^{4}2x\mathrm{d}x=16$$

This was much easier than finding the definite integral through its definition!

Evaluate

$${\int }_{-3}^3\sqrt{9-x^2}\mathrm{d}x.$$

Answer:

You are looking for the area below the function $f(x)=\sqrt{9-x^2}.$ By letting $y=f(x),$ you can write an equation and cancel the square root by squaring both sides, that is

$$\begin{array}{rl}y& =\sqrt{9-x^2}\\ y^2& =9-x^2,\end{array}$$

from where you can obtain the equation of the standard form of a circumference,

$$x^2+y^2=9.$$

Please note that this function is only the top half of the circle, as a whole circle would fail to be a function!

Fig. 5. The given function draws the upper part of a circle.

The radius of this circle is 3, so you can find its area by using the area of a circle formula, that is

$$\begin{array}{rl}{A}_C& =\pi r^2\\ & =\pi (3{)}^2\\ & =9\pi ,\end{array}$$

but since the area below the curve is half the area of the circle, you need to take half of this result. Therefore

$${\int }_{-3}^3\sqrt{9-x^2}\mathrm{d}x=\frac{9}{2}\pi .$$