Question:

Find the area under the curve represented by the function below, bounded by the x-axis and the vertical lines x = 1 and x = 5.

f(x) = x^2 - 3x + 2

Answer:

To find the area under the curve, we need to integrate the given function over the interval [1, 5]. Recall that the definite integral of a function f(x) over an interval [a, b] gives us the area between the curve and the x-axis over that interval.

Step 1: Evaluate the definite integral:

∫(1 to 5) (x^2 - 3x + 2) dx

We can break this integral down into three separate integrals:

∫(1 to 5) x^2 dx - ∫(1 to 5) 3x dx + ∫(1 to 5) 2 dx

Step 2: Integrate each term separately:

Using the power rule, we have:

∫(1 to 5) x^2 dx = [1/3 * x^3] evaluated from 1 to 5

= [1/3 * (5)^3] - [1/3 * (1)^3]

= (125/3) - (1/3)

= 124/3

For the second term:

∫(1 to 5) 3x dx = 3 * ∫(1 to 5) x dx

= 3 * [1/2 * x^2] evaluated from 1 to 5

= 3 * [1/2 * (5)^2] - [1/2 * (1)^2]

= 3 * (25/2) - (1/2)

= 75/2 - 1/2

= 74/2

= 37

For the third term:

∫(1 to 5) 2 dx = 2 * ∫(1 to 5) dx

= 2 * [x] evaluated from 1 to 5

= 2 * (5 - 1)

= 2 * 4

= 8

Now, let's sum up all the evaluated integrals:

∫(1 to 5) (x^2 - 3x + 2) dx = (124/3) - 37 + 8

= 124/3 - 37 + 8

= 124/3 - 9 + 24/3

= 124/3 - 9 + 8

= 124/3 - 1

= 41.33 - 1

= 40.33

Therefore, the area under the curve represented by the function f(x) = x^2 - 3x + 2, bounded by the x-axis and the vertical lines x = 1 and x = 5, is approximately 40.33 units squared.