Question Statement

Find the area bounded by the curve $f(x)=x^3-2x^2+1$ and the $x$-axis in the first quadrant bounded by the line $x=1.5$.

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Background and Explanation

This problem involves calculating the area between a cubic curve and the $x$-axis using definite integration. First, find the roots of the function to determine where the curve intersects the $x$-axis, then set up the appropriate definite integral over the interval where the function is non-negative in the first quadrant.

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Solution

Given the curve: $f(x)=x^3-2x^2+1$

Finding the Points of Intersection with the $x$-axis

At the $x$-axis, $f(x)=0$: $x^3-2x^2+1=0$

Finding Roots by Inspection

Take $x=1$: $(1{)}^3-2(1{)}^2+1=0$ $1-2+1=0$ $0=0$

Therefore, $x=1$ is a root by inspection.

Synthetic Division

Using synthetic division to factor out $(x-1)$ from the polynomial $x^3-2x^2+0x+1$:

The depressed equation is: $x^2-x-1=0$

Solving the Quadratic Equation

Here $a=1$, $b=-1$, $c=-1$. Using the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(1)(-1)}}{2\times 1}=\frac{1\pm \sqrt{1+4}}{2}$

$x=\frac{1\pm \sqrt{5}}{2}$

This yields: $x=\frac{1+\sqrt{5}}{2}\text{and}x=\frac{1-\sqrt{5}}{2}$

Since $\frac{1-\sqrt{5}}{2}$ is negative, we neglect it for the first quadrant.

Calculating the Area

Since the required area is in the first quadrant, and observing from the figure that the curve is above the $x$-axis from $x=0$ to $x=1$:

The area is given by:

$\text{Area}={\int }_0^1\left(x^3-2x^2+1\right)dx$

Breaking this into separate integrals: $\text{Area}={\int }_0^1{x}^{3}dx-2{\int }_0^1{x}^{2}dx+{\int }_0^{1}1dx$

Evaluating each term: $={\left[\frac{x^4}{4}\right]}_0^1-2{\left[\frac{x^3}{3}\right]}_0^1+{\left[x\right]}_0^1$

Substituting the limits: $=\left(\frac{1}{4}-0\right)-2\left(\frac{1}{3}-0\right)+(1-0)$

$=\frac{1}{4}-\frac{2}{3}+1$

Finding a common denominator (12): $=\frac{3-8+12}{12}$

$=\frac{7}{12}\text{ sq. unit}$

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Key Formulas or Methods Used

• Synthetic Division: For polynomial factorization when one root is known
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for solving $a{x}^2+bx+c=0$
• Definite Integration: Area $={\int }_a^{b}f(x)dx$ for calculating area under a curve
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$

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Summary of Steps

1. Set up the equation: Identify $f(x)=x^3-2x^2+1$ and find where $f(x)=0$
2. Find rational root: Verify $x=1$ is a root by substitution
3. Factor the polynomial: Use synthetic division to obtain the depressed equation $x^2-x-1=0$
4. Find remaining roots: Apply the quadratic formula to get $x=\frac{1\pm \sqrt{5}}{2}$
5. Determine integration limits: Identify that the curve is above the $x$-axis from $x=0$ to $x=1$ in the first quadrant
6. Set up the integral: Write $\text{Area}={\int }_0^1(x^3-2x^2+1)dx$
7. Evaluate: Compute the definite integral to obtain $\frac{7}{12}$ square units