Question Statement

Find the area bounded by the curve $y=x^3+1$, the $x$-axis and the line $x=1$.

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Background and Explanation

To find the area bounded by a curve and the $x$-axis, we evaluate the definite integral of the function over the interval where the curve lies above the axis. First, determine the left and right boundaries by finding where the curve intersects the $x$-axis and identifying any given vertical lines.

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Solution

Finding the Limits of Integration

We begin with the curve: $y=x^3+1$

The region is bounded on the right by the line $x=1$. To find the left boundary, we determine where the curve intersects the $x$-axis by setting $y=0$:

$\begin{array}{rl}{x}^3+1& =0\\ (x+1)\left(x^2+x+1\right)& =0\end{array}$

This yields: $x+1=0\Rightarrow x=-1$

For the quadratic factor $x^2+x+1=0$, we apply the quadratic formula:

$\begin{array}{rl}x& =\frac{-1\pm \sqrt{(1{)}^2-4(1)(1)}}{2}\\ & =\frac{-1\pm \sqrt{1-4}}{2}\\ & =\frac{-1\pm \sqrt{-3}}{2}\end{array}$

Since the discriminant is negative, these roots are imaginary. Therefore, the only real intersection with the $x$-axis occurs at $x=-1$.

The area is bounded between $x=-1$ and $x=1$.

Setting Up and Evaluating the Integral

The area is given by the definite integral:

$\begin{array}{rl}\text{Area}& ={\int }_{-1}^{1}ydx\\ & ={\int }_{-1}^1\left(x^3+1\right)dx\end{array}$

Using the linearity of integration, we split this into two separate integrals:

$\text{Area}={\int }_{-1}^1{x}^{3}dx+{\int }_{-1}^{1}1dx$

Evaluating the first integral using the power rule:

$\begin{array}{rl}{\int }_{-1}^1{x}^{3}dx& ={\left|\frac{x^4}{4}\right|}_{-1}^1\\ & =\frac{1}{4}\left((1{)}^4-(-1{)}^4\right)\\ & =\frac{1}{4}(1-1)\\ & =\frac{1}{4}(0)\\ & =0\end{array}$

Evaluating the second integral:

$\begin{array}{rl}{\int }_{-1}^{1}1dx& =|x{|}_{-1}^1\\ & =(1-(-1))\\ & =1+1\\ & =2\end{array}$

Combining both results:

$\begin{array}{rl}\text{Area}& =0+2\\ & =2\text{ sq. unit}\end{array}$

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Key Formulas or Methods Used

• Area under a curve: $A={\int }_a^{b}f(x)dx$ where $a$ and $b$ are the horizontal boundaries
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ (for $n\ne -1$)
• Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for solving $a{x}^2+bx+c=0$
• Linearity of integration: $\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

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Summary of Steps

1. Find x-intercept: Set $y=0$ and solve $x^3+1=0$ to find the left boundary at $x=-1$ (discarding imaginary roots)
2. Identify boundaries: The region extends from $x=-1$ (x-intercept) to $x=1$ (given line)
3. Set up integral: Write $\text{Area}={\int }_{-1}^1(x^3+1)dx$
4. Split the integral: Separate into ${\int }_{-1}^1{x}^{3}dx+{\int }_{-1}^{1}1dx$
5. Integrate: Apply the power rule to get ${\left[\frac{x^4}{4}\right]}_{-1}^1+[x{]}_{-1}^1$
6. Evaluate limits: Substitute bounds to obtain $\frac{1}{4}(1-1)+(1-(-1))=0+2$
7. Final answer: The area equals $2$ square units