3.7 Q-7

Question Statement

Find the area bounded by the curve $y = x^3 + 1$, the $x$-axis, and the line $x = 2$.

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Background and Explanation

In this problem, we are tasked with finding the area enclosed between the curve $y = x^3 + 1$, the $x$-axis, and the vertical line $x = 2$. To solve this, we need to:

1. Identify the points where the curve intersects the $x$-axis.
2. Set up the integral for the area between the curve and the $x$-axis within the given limits.

We can calculate the area by evaluating a definite integral of the function over the range from $x = -1$ (where the curve intersects the $x$-axis) to $x = 2$.

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Solution

Step 1: Find the x-intercepts

The curve intersects the $x$-axis when $y = 0$. Set the equation $y = x^3 + 1$ equal to zero:

$$x^3 + 1 = 0$$

We can factor this as:

$$(x + 1)(x^2 - x + 1) = 0$$

This gives two factors:

• $x + 1 = 0 \Rightarrow x = -1$
• $x^2 - x + 1 = 0$ has no real solutions because the discriminant $1 - 4 = -3$ is negative.

Thus, the curve only intersects the $x$-axis at $x = -1$.

Step 2: Set up the integral

The area we need to find is bounded by the curve from $x = -1$ to $x = 2$. The integral for the area is:

$$A = \int_{-1}^{2} (x^3 + 1) , dx$$

Step 3: Compute the integral

We now compute the integral of the function $x^3 + 1$:

• The integral of $x^3$ is $\frac{x^4}{4}$.
• The integral of $1$ is $x$.

So, the integral becomes:

$$A = \left[ \frac{x^4}{4} + x \right]_{-1}^{2}$$

Step 4: Evaluate the definite integral

Now, evaluate the expression at the limits $x = 2$ and $x = -1$:

For $x = 2$:

$$\frac{(2)^4}{4} + 2 = \frac{16}{4} + 2 = 4 + 2 = 6$$

For $x = -1$:

$$\frac{(-1)^4}{4} + (-1) = \frac{1}{4} - 1 = \frac{1}{4} - \frac{4}{4} = -\frac{3}{4}$$

Step 5: Final calculation

Now, subtract the result at $x = -1$ from the result at $x = 2$:

$$A = 6 - \left( -\frac{3}{4} \right) = 6 + \frac{3}{4} = \frac{24}{4} + \frac{3}{4} = \frac{27}{4}$$

Thus, the area is:

$$A = \frac{27}{4} \text{ square units}$$

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Key Formulas or Methods Used

• Definite Integral: The area under a curve from $x = a$ to $x = b$ is given by:

$$A = \int_a^b f(x) , dx$$

• Power Rule of Integration:

$$\int x^n , dx = \frac{x^{n+1}}{n+1}$$

• Factoring: Factoring the cubic equation $x^3 + 1 = 0$ to find the $x$-intercepts.

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Summary of Steps

1. Set $y = x^3 + 1$ and solve for $x$ to find the $x$-intercepts: $x = -1$.
2. Set up the integral to compute the area: $A = \int_{-1}^{2} (x^3 + 1) , dx$
3. Compute the integral: $A = \left[ \frac{x^4}{4} + x \right]_{-1}^{2}$
4. Evaluate the integral at the limits: $A = 6 - \left( -\frac{3}{4} \right) = \frac{27}{4}$
5. The final area is: $A = \frac{27}{4} \text{ square units}$