Area under a curve

What you are looking at

A definite integral adds up $f(x)$ times a small width across $[a, b]$. The visualiser draws that sum as strips: left, right or midpoint rectangles, or trapezia joining neighbouring points on the curve. The readout gives the strip total, the exact integral and the error between them, so you can watch the error shrink as the strips get narrower. With the trapezium rule, the working shows $\dfrac{h}{2}\big(y_0 + 2(y_1 + \dots + y_{n-1}) + y_n\big)$ with the current strip width $h$.

Strips below the axis are drawn red and count as negative, because $f(x)$ is negative there. That is why the integral and the area can be different numbers. Take $f(x) = x^3$ from $-1$ to $1$: the red strips cancel the blue ones and $\int_{-1}^{1} x^3\,dx = 0$, while the area between the curve and the axis is $\int_{-1}^{1} |x^3|\,dx = \dfrac{1}{2}$. Both values are shown for every function and interval you choose.

Try this

1. 1Keep $f(x) = x^2$ on $[0, 2]$ with left rectangles and push the strips slider up from 6: the total rises towards $\int_0^2 x^2\,dx = \dfrac{8}{3} \approx 2.667$ and the error shrinks.
2. 2Choose $x^3$ and set $a = -1$, $b = 1$: the red strips cancel the blue, so the integral is 0 while the area is 0.5.
3. 3On $x^2$ with 6 strips, switch between Trapezium and Midpoint: the trapezium rule overestimates by about 0.037 and the midpoint rule underestimates by about 0.019, because the curve bends upwards.
4. 4Drag $b$ to the left of $a$: the integral changes sign, but the area stays the same.

Mistakes students make

• Integrating straight across a root to find an area. The integral counts the region below the axis as negative, so split the interval at the roots and add the sizes of the parts, or integrate $|f(x)|$.
• Thinking left rectangles always underestimate. They do only when $f$ is increasing on $[a, b]$; for a decreasing function they overestimate.
• Miscounting ordinates in the trapezium rule. With $n$ strips there are $n + 1$ values $y_0, \dots, y_n$, the width is $h = \dfrac{b-a}{n}$, and only the two end values are not doubled.

Where it is taught

• IB Maths AA SL/HL 5.5 · 5.11
• AP Calculus AB/BC Unit 6
• Edexcel A-Level Pure Year 1 · Integration (areas under curves)
• Edexcel A-Level Pure Year 2 · Integration (the trapezium rule)
• Greek Lykeio year 3 (Γ΄ Λυκείου) · §3.4 · §3.7