Question

In Exercises \(43-48,\) use the limit process to find the area of the region between the graph of the function and the \(x\) -axis over the given interval. Sketch the region. $$ y=-2 x+3, \quad[0,1] $$

Short answer

The area of the region between the graph of the function \(y = -2x + 3\) and the \(x\) -axis over the interval [0,1] is 2 square units.

Step-by-step solution

Sketch the Region

Start by graphing this linear function on the interval [0,1]. This line intersects the x-axis around 1.5, so on [0,1] it lies completely above the x-axis.

Setup the Integral

Since we are finding the area between the function and the x-axis over the interval [0,1], setup the integral as follows: \(\int_{0}^{1} (-2x + 3)dx\).

Evaluate the Integral

To evaluate this definite integral, first find the antiderivative of the integrand. An antiderivative of \(-2x + 3\) is \(-x^2 + 3x\). Then, apply the Second Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at 0 from the antiderivative evaluated at 1. This gives \((-(1)^2 + 3 * (1)) - (-(0)^2 + 3 * (0)) = 2\).