Finding the Area Bounded by a Curve

Area needs absolute value, not a signed integral

A definite integral gives signedarea: positive above the x-axis, negative below it. To find the actual (unsigned) area enclosed by a curve that dips below the axis, integrate over each piece where the sign doesn't change, and add the absolute values together.

Worked example: a full period of cosine

Find the area bounded by $y=\cos x$ between $x=0$ and $x=2\pi$.

Solution: $\cos x \ge 0$ on $[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},2\pi]$ and $\cos x \le 0$ on $[\frac{\pi}{2},\frac{3\pi}{2}]$. By symmetry, the total area is 4 times the area of one quarter-period:

Worked example: deriving the area of a circle

Find the area of the circle $x^2+y^2=a^2$ by integration.

Solution: The circle's upper half is $y=\sqrt{a^2-x^2}$. By symmetry, the full circle's area is 4 times the area in the first quadrant:

At $x=a$ the square-root term vanishes and $\sin^{-1}(1) = \frac{\pi}{2}$, while both terms vanish at $x=0$: