Question

Approximate the following integrals using either the midpoint rule, trapezoidal rule, or Simpson's rule as indicated. (Round answers to three decimal places.) ${\int }_0^3\sqrt{4+x^3}dx;$ trapezoidal rule; $n=6$

Short answer

The integral is approximately 9.189.

Step-by-step solution

Identify the interval and calculate the step size

The integral is over the interval $$a,b]=[0,3]$$. With $n=6$ subintervals, the step size $h$ is given by $$h=\frac{b-a}{n}=\frac{3-0}{6}=0.5.$$

Calculate the x-values at each step

Subdivide the interval $$0,3$$ into 6 subintervals of length 0.5. The x-values are: $$x_0=0,x_1=0.5,x_2=1.0,x_3=1.5,x_4=2.0,x_5=2.5,x_6=3.0.$$

Evaluate the function at each x-value

Evaluate $f(x)=\sqrt{4+x^3}$ at each of the x-values: $$f(x_0)=\sqrt{4+0^3}=2.$$$$f(x_1)=\sqrt{4+(0.5{)}^3}\approx 2.013.$$$$f(x_2)=\sqrt{4+1^3}=\sqrt{5}\approx 2.236.$$$$f(x_3)=\sqrt{4+(1.5{)}^3}\approx 2.756.$$$$f(x_4)=\sqrt{4+2^3}=\sqrt{12}\approx 3.464.$$$$f(x_5)=\sqrt{4+(2.5{)}^3}\approx 4.125.$$$$f(x_6)=\sqrt{4+3^3}=\sqrt{31}\approx 5.568.$$

Apply the Trapezoidal Rule formula

The Trapezoidal Rule formula for an integral ${\int }_a^{b}f(x)dx$ is given by $$\frac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+2f(x_4)+2f(x_5)+f(x_6)].$$Plug in the values: $$\text{Integral}\approx \frac{0.5}{2}[2+2(2.013)+2(2.236)+2(2.756)+2(3.464)+2(4.125)+5.568].$$

Calculate the approximation

Perform the calculations: $$\text{Integral}\approx 0.25\times (2+4.026+4.472+5.512+6.928+8.25+5.568).$$Adding these up gives: $$\text{Integral}\approx 0.25\times 36.756\approx 9.189.$$

Round the result

Round the result to three decimal points, the approximate value of the integral is $$9.189.$$

Key concepts

Trapezoidal Rule

The trapezoidal rule is a technique used for numerical integration, allowing you to approximate definite integrals. It's particularly useful when an antiderivative of the function is difficult or impossible to find. The basic idea is to replace the actual curve of the function with straight line segments, forming a series of trapezoids. By calculating the sum of these trapezoids' areas, you can estimate the total area under the curve over a specified interval.
The formula for the trapezoidal rule to approximate the integral of a function $f(x)$ over an interval $[a,b]$ is given by:

• ${\int }_a^{b}f(x)\approx \frac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+\cdots +2f(x_{n-1})+f(x_n)]$

where $h$ is the step size, and $x_0,x_1,\dots ,x_n$ are the division points of the interval. The endpoints $f(x_0)$ and $f(x_n)$ are not doubled because they are only counted once. This method is straightforward and easy to apply, making it a popular choice for simple midpoint calculations.

Step Size Calculation

In the context of numerical integration, the step size, usually denoted as $h$, is a critical factor. It determines the width of each small subinterval into which you divide the overall integration span. Calculating this step size is essential, as it greatly affects the accuracy of the integration result.
To calculate $h$, you use the formula:

• $h=\frac{b-a}{n}$

where $a$ and $b$ are the bounds of the integral, and $n$ is the number of subintervals. Each subinterval has a length of $h$, and a smaller value of $h$ typically results in more precise approximations, as the subdivided sections better conform to the curve's shape. However, increasing $n$ also requires more computational work, so there's a trade-off between accuracy and computational efficiency. This balance is a fundamental aspect of numerical methods.

Function Evaluation

Once the interval is divided and the step size is calculated in numerical integration, function evaluation becomes the next step. This involves determining the value of the function at specific points along the interval. For the trapezoidal rule, as well as other numerical methods, these evaluations are crucial parts of the approximation process.

• At each division point $x_i$, you will evaluate the function $f(x_i)$.
• The results of these evaluations are then integrated into the trapezoidal formula, which multiplies them by the assigned weights (usually 1 for the endpoints and 2 for the intermediate points).

These calculated values of the function at selected $x$-values help translate the function into discrete points that can approximate the area under the curve. The more points evaluated, generally the more accurate the approximation becomes, particularly if the function exhibits significant changes across the interval.