Question

To evaluate the definite integral.

Short answer

The evaluation of the definite integral is \(\frac{{\ln (16)}}{5}\).

Step-by-step solution

Given data

The definite integral function is \(\int_0^3 {\frac{{dx}}{{5x + 1}}} \). The region lies between \(x = 0\) and \(x = 3\).

Concept used of Definite Integral and Differentiation

The definite integral of a real-valued function\(f(x)\)with respect to a real variable\(x\)on an interval\((a,b)\)expressed as\(\int_a^b f (x)dx = F(b) - F(a)\).

Differentiation

Let\(y = f(x)\)be a function of\(x\). Then, the rate of change of "\(y\)" per unit change in "\(x\)" is given by\(dy/dx\).

Differentiate the given function

Consider

\(u = 5x + 1\) …..(1)

Differentiate both sides of the Equation (1).

\(\begin{aligned}{l}du &= 5dx\\dx &= \frac{1}{5}du\end{aligned}\)

Calculate the lower limit value of \(u\) by the use of Equation (1).

Substitute 0 for \(x\) in Equation (1).

\(u = 5(0) + 1 = 1\)

Calculate the upper limit value of \(u\) by the use of Equation (1).

Substitute 1 for \(x\) in Equation (1).

\(u = 5(3) + 1 = 16\)

The definite integral function is,

\(\int_0^3 {\frac{{dx}}{{5x + 1}}} \) …..(2)

Apply lower and upper limits

Apply the lower and upper limit for \(u\) in Equation (2).

Substitute \(u\) for \((5x + 1)\) and \(\left( {\frac{1}{5}du} \right)\) for \(dx\)

\(\begin{aligned}{c}\int_0^3 {\frac{{dx}}{{5x + 1}}} = \int_1^{16} {\frac{1}{u}} \left( {\frac{1}{5}du} \right)\\ = \frac{1}{5}\int_1^{16} {\frac{1}{u}} du\end{aligned}\) ……(3)

Integrate Equation (3)