Quick questions on Area under and between curves: VCE Math Methods Unit 4

If you compute $\int_{a}^{b} f \, dx$ without splitting and the curve crosses zero, positive and negative regions cancel. The arithmetic result is the net signed area, not the geometric area.

1. Find the intersection points. Solve $f(x) = g(x)$. These set the integration limits if the question asks for "the enclosed area". 2.

The curves may cross more than once in the interval of interest. Each sub-interval needs its own setup with the correct top and bottom.

If $g(x) = 0$ (the $x$-axis), the formula reduces to $\int [f(x) - 0] \, dx = \int f(x) \, dx$ when $f \geq 0$. This recovers the single-curve case.

Area between $y = x^3 - 4x$ and the $x$-axis on $[-2, 2]$.

Area enclosed between $y = x^2$ and $y = x + 2$.

Area enclosed between $y = \sin(x)$ and $y = \cos(x)$ on $[0, 2\pi]$ requires finding all intersections and splitting accordingly. The intersections on $[0, 2\pi]$ are at $x = \pi/4$ and $x = 5\pi/4$.

Computing $\int_{a}^{b} f \, dx$ for a function that crosses zero and reporting the result as "area" gives the wrong (smaller, possibly zero) answer.

Picking the wrong top yields a negative answer. The fix: test a value inside the interval and pick the larger.

If the question asks for the "enclosed area" between two curves without giving limits, the intersection points provide the limits.

"The value of $\int f \, dx$" is the signed area; "the area" is geometric (non-negative). Read the question carefully.

$F(b) - F(a)$ requires brackets when $F$ has multiple terms. Sign errors here cascade.

Taking the absolute value of the result of a single integral is wrong if the integrand changes sign on the interval. The absolute value must go on each sub-integral (or you must integrate $|f|$ piecewise).