Find the area bounded by the curve y = x² + x + 1

Find the area bounded by the curve $y = x^2 + x + 1$, the lines $x = 0$, $x = 1$, and $y = 0$.

We will find the area using a straightforward definite integral.

Step 1: Understand the region

We need to find the area of the region bounded by:

• The curve: $y = x^2 + x + 1$
• Left boundary: $x = 0$
• Right boundary: $x = 1$
• Bottom boundary: $y = 0$ (the $x$-axis)

Step 2: Verify that the curve is above the $x$-axis

For $x \in [0, 1]$:

\[y = x^2 + x + 1 \geq 0 + 0 + 1 = 1 > 0\]

So the curve is entirely above the $x$-axis on this interval. ✓

Step 3: Set up the integral

The area is:

\[A = \int_0^1 (x^2 + x + 1) \, dx\]

Step 4: Calculate the antiderivative

\[\int (x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C\]

Step 5: Evaluate the definite integral

\[A = \left[\frac{x^3}{3} + \frac{x^2}{2} + x\right]_0^1\]

At $x = 1$:

\[\frac{1^3}{3} + \frac{1^2}{2} + 1 = \frac{1}{3} + \frac{1}{2} + 1 = \frac{2 + 3 + 6}{6} = \frac{11}{6}\]

At $x = 0$:

\[\frac{0^3}{3} + \frac{0^2}{2} + 0 = 0\]

Therefore:

\[A = \frac{11}{6} - 0 = \frac{11}{6}\]

\[\boxed{A = \frac{11}{6} \text{ square units}}\]

Step 6: Numerical value

\[A = \frac{11}{6} \approx 1.833 \text{ square units}\]

Verification

Let's verify by checking the calculation:

\[\frac{1}{3} + \frac{1}{2} + 1 = \frac{2}{6} + \frac{3}{6} + \frac{6}{6} = \frac{11}{6} \quad \checkmark\]

Summary

The area bounded by the curve $y = x^2 + x + 1$, the lines $x = 0$, $x = 1$, and $y = 0$ is:

\[\boxed{A = \frac{11}{6} \approx 1.833 \text{ square units}}\]

Key steps:

1. Identify the region (curve above $x$-axis on $[0,1]$)
2. Set up integral: $A = \int_0^1 (x^2 + x + 1) \, dx$
3. Find antiderivative: $\frac{x^3}{3} + \frac{x^2}{2} + x$
4. Evaluate: $\frac{11}{6} - 0 = \frac{11}{6}$