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Calculus

The Definite Integral

As discussed in the previous section, the process of Riemann sums leads to the definition of the definite integral. We can denote the definite integral as

\[\int_{a}^{b} f(x)dx = \lim_{n\to\infty} \sum\limits_{i=0}^{n-1} f(x_i)\Delta x\]

To evaluate a definite integral, we can express it as the difference of two antiderivatives at their end points.

\[\int_{a}^{b} f(x)dx = F(b) - F(a)\]

where \(F'(x)=f(x)\).

Example: Evaluate the definite integral \[\int_{1}^{4} \frac{y+4}{\sqrt{y}}dy\]

Solution:

First, we want to evaluate the integral 

\[\int \frac{y+4}{\sqrt{y}}dy\]

This integral can be simplified by splitting the fraction and simplifying roots into fractional exponents

\begin{align} &\int \frac{y+4}{\sqrt{y}}dy \\=&\int \left(\frac{y}{\sqrt{y}} + \frac{4}{\sqrt{y}}\right)dy \\ =&\int \left( y^{\frac{1}{2}} + 4y^{-\frac{1}{2}} \right)dy \end{align}

Now we can evaluate the integral with constant c

\begin{align} =& \frac{y^{\frac{3}{2}}}{\frac{3}{2}} + \frac{4y^{\frac{1}{2}}}{\frac{1}{2}} + c \\ =& \frac{2y^{\frac{3}{2}}}{3} + 8y^{\frac{1}{2}} + c \end{align}

To evaluate the definite integral, we sub in the end points \([1,4]\)

\begin{align} F(4) - F(1) =& \left[\frac{2y^{\frac{3}{2}}}{3} + 8y^{\frac{1}{2}} + c \right]_{1}^{4} \\ =& \left[\frac{2(\textbf{4})^{\frac{3}{2}}}{3} + 8(\textbf{4})^{\frac{1}{2}} + c \right] - \left[\frac{2(\textbf{1})^{\frac{3}{2}}}{3} + 8(\textbf{1})^{\frac{1}{2}} + c \right] \\ =& \left[ \frac{64}{3}\right] - \left[ \frac{26}{3} \right] \\ =& \frac{38}{3} \end{align}

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