Question

In Exercises \(5-8,\) the integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{0}^{4}\left[(x+1)-\frac{x}{2}\right] d x $$

Short answer

The area between the functions \(x+1\) and \(\frac{x}{2}\) from \(x=0\) to \(x=4\) is 8.

Step-by-step solution

Sketch the graphs

First the two functions, \(x+1\) and \(\frac{x}{2}\), need to be plotted on the same graph. The function \(x+1\) is a linear function with a slope of one and a y-intercept of one. The function \(\frac{x}{2}\) is a linear function with a slope of 0.5 and goes through the origin.

Shade the region representing the integral

The region represented by the integral is the area between the two functions from \(x=0\) to \(x=4\). So the region between the two functions and between \(x=0\) and \(x=4\) is shaded.

Calculate the integral

The integral is calculated as \(\int_{0}^{4}\left[(x+1)-\frac{x}{2}\right] d x\). First break down the integral into two separate integrals \(\int_{0}^{4}(x+1) d x - \int_{0}^{4}\frac{x}{2} d x\). Then calculate each integral individually. The integral of \(x+1\) from \(0\) to \(4\) is \(\frac{1}{2}x^{2}+x\) evaluated from \(0\) to \(4\) which gives 12. The integral of \(\frac{x}{2}\) from \(0\) to \(4\) is \(\frac{1}{2} * \frac{1}{2}x^{2}\) evaluated from \(0\) to \(4\) which gives 4. Subtract these two results to get the value of the original integral which is 12 - 4 = 8.

Interpret the result

The value of the integral, which is 8, represents the signed area between the functions \(x+1\) and \(\frac{x}{2}\) over the interval \([0,4]\). In this context, it is the area above the line \(\frac{x}{2}\) and below the line \(x+1\) from \(x=0\) to \(x=4\).