Math Problem Statement

Find the area of the region enclosed by the graph y=x^2-4x+3 and y=0 between x=0 and x=4.

Solution

We are tasked with finding the area of the region enclosed by the graph y=x24x+3y = x^2 - 4x + 3 and the line y=0y = 0 between x=0x = 0 and x=4x = 4.

Step 1: Find the points where the curve intersects the x-axis

To determine the region enclosed by the graph and the x-axis, we need to find the points where the curve intersects the x-axis (i.e., where y=0y = 0). Set the equation y=x24x+3=0y = x^2 - 4x + 3 = 0 and solve for xx:

x24x+3=0x^2 - 4x + 3 = 0

We can factor the quadratic equation:

(x1)(x3)=0(x - 1)(x - 3) = 0

Thus, the roots are:

x=1andx=3x = 1 \quad \text{and} \quad x = 3

Step 2: Set up the integral

Now, we need to find the area between the curve and the x-axis from x=0x = 0 to x=4x = 4. The area is given by the integral of the function y=x24x+3y = x^2 - 4x + 3 over this interval:

Area=04(x24x+3)dx\text{Area} = \int_0^4 (x^2 - 4x + 3) \, dx

Step 3: Compute the integral

Now, let's compute the integral:

(x24x+3)dx=x332x2+3x\int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x

We will now evaluate this from x=0x = 0 to x=4x = 4:

Area=[x332x2+3x]04\text{Area} = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_0^4

First, substitute x=4x = 4:

4332(42)+3(4)=6432(16)+12=64332+12=64320\frac{4^3}{3} - 2(4^2) + 3(4) = \frac{64}{3} - 2(16) + 12 = \frac{64}{3} - 32 + 12 = \frac{64}{3} - 20

Next, substitute x=0x = 0:

0332(02)+3(0)=0\frac{0^3}{3} - 2(0^2) + 3(0) = 0

Thus, the area is:

Area=64320=643603=43\text{Area} = \frac{64}{3} - 20 = \frac{64}{3} - \frac{60}{3} = \frac{4}{3}

Step 4: Interpret the result

The area of the region enclosed by the graph y=x24x+3y = x^2 - 4x + 3 and the x-axis between x=0x = 0 and x=4x = 4 is 43\frac{4}{3}.


Would you like more details on any part of the process? Let me know if you have further questions!

Here are some related questions you might find interesting:

  1. How do you find the area between two curves?
  2. What is the significance of the roots in finding the enclosed area?
  3. Can you apply this method to other polynomial functions?
  4. What happens if the function does not intersect the x-axis within the given interval?
  5. How do you handle cases where the graph is below the x-axis in finding the area?

Tip: Always ensure that the function is properly bounded by the region of interest before setting up your integral!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Definite Integration
Area under a Curve

Formulas

Quadratic equation formula ax^2 + bx + c = 0
Definite integral formula ∫[a,b] f(x) dx

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 10-12