Area and slope undo each other
Before this sheet: The slope of a curve and Area under a curve.
Tonight's idea
Let \(A(x)\) be the area under \(y = t^2\), from \(t = 0\) to \(t = x\), where \(x \ge 0\). The letter \(t\) runs along the graph; \(x\) marks the movable end of the region.
Push that end a little to the right, by \(h\). Predict: what sets the extra area's size — the whole curve, or just its height near the moving edge?
The extra area is \(A(x+h)-A(x)\). Divide by its width \(h\): that is the strip's average height, and also the slope between two points on the graph of \(A\). For \(h > 0\),
\[ x^2 \le \frac{A(x+h)-A(x)}{h} \le (x+h)^2. \]As \(h\) shrinks to zero, both bounds tend to \(x^2\). Moving the edge left gives the same limit when \(x > 0\). So \(A'(x)=x^2\): the slope of accumulated area is the curve's height.
For a continuous curve \(f\), accumulated area has derivative \(f\). Below the axis, count area negatively. This is the fundamental theorem of calculus.
Watch one: find an area by checking a slope
The area sheet found \(A(1)=\tfrac13\). Stretching that region to end at \(x\) multiplies widths by \(x\) and heights by \(x^2\), so \(A(x)=\tfrac{x^3}{3}\). Check: its derivative is \(x^2\), and \(A(0)=0\).
From 1 to 2, subtract the area you don't want:
\[ A(2)-A(1)=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}. \]More generally, if \(F'=f\), then \(F\) and the accumulated area have the same slope, so they differ only by a constant. Subtracting cancels it: the signed area from \(a\) to \(b\) is \(F(b)-F(a)\). Find an area by finding a function with the right derivative.
Your turn
- Under \(y=2t\), from 0 to \(x\), the region is a triangle. Find \(A(x)\) using its base and height, then check that \(A'(x)=2x\).
- Find the area under \(y=t^2\) from 2 to 3. Why would using \(A(3)\) alone give too much?
- A car's speed is \(2t+3\) metres per second, from \(t=0\) to \(t=4\) seconds. Find its distance using a rectangle and a triangle. Check by finding a function whose derivative is \(2t+3\).
- A curve stays on or above the axis. Could its accumulated-area graph ever slope downwards? Explain.
Stretch: home again, but how far?
A car's velocity is \(v(t)=2t-4\) metres per second for \(0 \le t \le 4\). Negative velocity means travelling backwards. Sketch the velocity graph. Use signed area to find its final displacement from the start, then find the total distance it travelled. Why are the answers different?
Answers
- The base is \(x\), the height \(2x\): \(A(x)=\tfrac12 x(2x)=x^2\), and \(A'(x)=2x\).
- \(A(3)-A(2)=\tfrac{27}{3}-\tfrac{8}{3}=\tfrac{19}{3}\). \(A(3)\) includes the unwanted region from 0 to 2.
- The rectangle has area \(4 \times 3=12\); the triangle has area \(\tfrac12 \times 4 \times 8=16\). Total: 28 metres. \(D(t)=t^2+3t\) has the right derivative; \(D(4)-D(0)=28\).
- No. Its slope is the curve's height, which is nonnegative. It can be flat where the height is zero.
Stretch. Velocity is zero at \(t=2\). The triangle below the axis has signed area \(-4\) metres; the one above has area \(+4\). Displacement: 0 metres — back at the start. Distance: \(4+4=8\) metres. Checking with \(F(t)=t^2-4t\) gives \(F(4)-F(0)=0\); signed area records direction as well as amount.