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 \( \Delta x \) :

\[ \begin{align} \Delta x &= \frac{b-a}{N} \\ &= \frac{5-0}{N} \\ &= \frac{5}{N}. \end{align} \]

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

\[ \int_0^5 x^2\,\mathrm{d}x = \lim_{N\rightarrow \infty} \left[ \sum_{i=1}^{N}\left( x_i^* \right)^2 \left( \frac{5}{N}\right) \right] .\]

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

\[ \begin{align} x_i^* &= a+i\Delta x \\ &= i\frac{5}{N}, \end{align} \]

which can be substituted back into the definite integral

\[ \int_0^5 x^2\,\mathrm{d}x = \lim_{N\rightarrow \infty} \left[ \sum_{i=1}^{N}\left( i\frac{5}{N} \right)^2 \left( \frac{5}{N}\right) \right] .\]

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

\[\begin{align} \int_0^5 x^2\,\mathrm{d}x &= \lim_{N\rightarrow \infty} \left[ \sum_{i=1}^{N}\left( i\frac{5}{N} \right)^2 \left( \frac{5}{N}\right) \right] \\ &= \lim_{N\rightarrow \infty} \left[ \left( \frac{5}{N}\right)^3 \sum_{i=1}^{N}i^2 \right]. \end{align}\]

You can use the formula

\[\sum_{i=1}^{N}i^2=\frac{N(N+1)(2N+1)}{6}\]

to evaluate the sum, so

\[ \int_0^5 x^2 \,\mathrm{d}x = \lim_{N\rightarrow \infty} \left[ \left(\frac{5}{N}\right)^3 \left(\frac{N(N+1)(2N+1)}{6}\right) \right]. \]

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

\[ \int_0^5 x^2\,\mathrm{d}x = \lim_{N\rightarrow \infty} \left( \frac{125}{3}+\frac{125}{2N}+\frac{125}{6N^2} \right). \]

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

\[ \begin{align} \int_0^5 x^2 \, \mathrm{d}x &= \left( \frac{1}{3}(5)^3 \right) - \left( \frac{1}{3}(0)^3 \right) \\ &= \frac{125}{3}. \end{align} \]

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

\[ \begin{align} A &= \frac{bh}{2} \\ &= \frac{4(8)}{2} \\ &= 16. \end{align} \]

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

\[ \begin{align} y &= \sqrt{9-x^2} \\ y^2 &=9-x^2, \end{align}\]

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

\[\begin{align} A_C &= \pi r^2 \\ &= \pi(3)^2 \\ &= 9\pi, \end{align} \]

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

\[\int_{-3}^3\sqrt{9-x^2} \, \mathrm{d}x = \frac{9}{2}\pi. \]