Integration – Finding Area

The area under a curve between \(x=a\) and \(x=b\) is:

No +C (definite integrals do not need a constant).

Rewrite roots and fractions as powers:

• \(\sqrt{x} = x^{1/2}\)
• \(\frac{1}{x^3} = x^{-3}\)
• \(\sqrt[3]{x} = x^{1/3}\)

Find the area bounded by \(y = x^2\), the x-axis and the lines \(x=0\) and \(x=3\).

\[ \text{Area} = \int_0^3 x^2\, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} = 9. \] Area = 9 units².

Find the area under \(y=\sqrt{x}\) from \(x=1\) to \(x=4\).

Rewrite: \( \sqrt{x} = x^{1/2} \) \[ \text{Area} = \int_1^4 x^{1/2} dx = \left[\frac{2}{3}x^{3/2}\right]_1^4 \] \[ = \frac{2}{3}\left(4^{3/2} - 1^{3/2}\right) = \frac{2}{3}(8 - 1) = \frac{14}{3}. \] Area = \( \frac{14}{3} \) units².

Find the area bounded by \(y = x - 3\) from \(x=0\) to \(x=3\).

The curve lies below the x-axis for all \(x \in [0,3]\).

\[ \text{Area} = -\int_0^3 (x - 3)\, dx = -\left[\frac{x^2}{2} - 3x\right]_0^3 \] \[ = -\left(\frac{9}{2} - 9\right) = -\left(-\frac{9}{2}\right) = \frac{9}{2}. \] Area = \( \frac{9}{2} \) units².

Find the area between \(y = x^2 - 4x\) and the x-axis.

First solve where the curve crosses the x-axis:

\[ x^2 - 4x = x(x-4) = 0 \Rightarrow x=0,4. \]

\[ \text{Area} = \int_0^4 (4x - x^2)\, dx \] (We write it positive.) \[ = \left[2x^2 - \frac{x^3}{3}\right]_0^4 = \left(32 - \frac{64}{3}\right) = \frac{32}{3}. \] Area = \( \frac{32}{3} \) units².

• Always rewrite roots and reciprocals as powers.
• Never add +C when finding areas.
• If the curve is below the x-axis, make the area positive.
• When the curve crosses the x-axis, split the integral at the intercepts.
• Show all steps, especially solving for intercepts.