Get Ready: Module 4 — Integration & the Fundamental Theorem
Everyone starts somewhere. Before we measure the area under a curve, we'll sharpen the four tools that make it possible. This page rebuilds the handful of skills Module 4 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.
Module 4 runs the machinery of derivatives in reverse to recover areas and totals. If reading a derivative backward, the basic function shapes, sigma notation \(\sum\), or the area of a rectangle feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.
Skills to build first
Four short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.
Reading a derivative backward (antiderivatives)
What it is: in Modules 2–3 you learned that the derivative of \(x^3\) is \(3x^2\). An antiderivative just runs that arrow backward: "what function has this as its derivative?" The reverse power rule says \(\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) (add one to the power, divide by the new power). Why you need it: the Fundamental Theorem turns every area problem into "find an antiderivative, then subtract." If you can differentiate, you can learn to undo it.
- Add one to the power. The power is \(2\), so the new power is \(2+1=3\): we expect an \(x^3\) term.
- Divide by the new power. \(\dfrac{x^{3}}{3}\).
- Add the constant. Because the derivative of any constant is \(0\), write \(F(x)=\dfrac{x^3}{3} + C\). The \(+C\) is not optional for an indefinite integral.
- Check by differentiating. \(\dfrac{d}{dx}\!\left(\dfrac{x^3}{3}+C\right)=x^2\). ✓ Back where we started.
Try it 1 — Find an antiderivative of \(f(x)=x^4\).
Add one to the power (\(4\to 5\)) and divide: \(\displaystyle\int x^4\,dx = \dfrac{x^5}{5} + C\). Check: the derivative of \(\tfrac{x^5}{5}\) is \(x^4\). ✓
Try it 2 — Find an antiderivative of \(f(x)=6x\).
\(\displaystyle\int 6x\,dx = 6\cdot\dfrac{x^2}{2} + C = 3x^2 + C\). Check: the derivative of \(3x^2\) is \(6x\). ✓
The basic function families & their shapes
What it is: knowing, at a glance, the graph of the parents — the line \(y=mx+b\), the parabola \(y=x^2\), the square root \(y=\sqrt{x}\), and a sine wave \(y=\sin x\). Why you need it: to shade "the area under \(f\)" you first have to picture \(f\). Recognizing the family also tells you whether the curve dips below the \(x\)-axis — which is exactly where the integral's signed (net) area comes in.
- Name the family. A power of \(x^2\) is a parabola — a U-shape opening up, vertex at the origin.
- Plot a few anchor points. \(f(0)=0,\ f(1)=1,\ f(2)=4,\ f(3)=9\).
- Note it stays above the axis on \([0,3]\): every output is \(\ge 0\), so the area there is all positive.
- Read the trend. The curve is increasing and getting steeper — useful when you predict whether a left or right Riemann sum over- or under-shoots.
Try it 1 — On \([0,\pi]\), is \(f(x)=\sin x\) ever negative?
No. On \([0,\pi]\) the sine curve rises from \(0\) up to \(1\) (at \(x=\tfrac{\pi}{2}\)) and back to \(0\) — all outputs are \(\ge 0\), so the area is entirely above the axis. (It would go negative on \([\pi,2\pi]\).)
Try it 2 — Where does \(f(x)=x^3-3x\) cross the \(x\)-axis?
Set \(x^3-3x=0\Rightarrow x(x^2-3)=0\), so \(x=0,\ \pm\sqrt{3}\approx\pm1.73\). Between those crossings the cubic swaps sign — so part of its area lies below the axis (negative signed area).
Summation (sigma) notation
What it is: a compact way to write "add these up." \(\displaystyle\sum_{i=1}^{n} a_i\) means "start the counter \(i\) at \(1\), plug it into \(a_i\), and add the results up to \(i=n\)." You met this in Pre-Calculus Module 5. Why you need it: a Riemann sum is a sigma sum — \(\displaystyle\sum_{i=1}^{n} f(x_i)\,\Delta x\) — so reading the \(\Sigma\) fluently is reading the definition of the integral.
- List the terms. Let \(i\) run \(1,2,3,4\) and square each: \(1^2,\ 2^2,\ 3^2,\ 4^2\).
- Compute each. \(1,\ 4,\ 9,\ 16\).
- Add them up. \(1+4+9+16 = 30\).
That's the whole idea: \(\Sigma\) is just an instruction to loop a counter and total the outputs — exactly what a Riemann sum does with rectangle areas.
Try it 1 — Evaluate \(\displaystyle\sum_{i=1}^{5} 2i\).
Terms: \(2(1),2(2),2(3),2(4),2(5) = 2,4,6,8,10\). Sum \(= 30\).
Try it 2 — Write "the sum of \(f(x_1)\Delta x\) through \(f(x_n)\Delta x\)" in sigma notation.
\(\displaystyle\sum_{i=1}^{n} f(x_i)\,\Delta x\). This is the Riemann sum — the very thing that becomes \(\int_a^b f(x)\,dx\) as \(n\to\infty\).
Areas of basic shapes
What it is: the area of a rectangle is base × height; the area of a triangle is \(\tfrac12\,bh\); the area of a trapezoid is \(\tfrac12(b_1+b_2)h\) (average of the two parallel sides, times the width between them). Why you need it: a Riemann rectangle's area is literally width \(\times\) height, and the trapezoid method uses the trapezoid formula. Some integrals are just a triangle or trapezoid you can find with geometry — a great way to check the calculus.
- Find the two end heights. At \(x=0\), \(y=1\); at \(x=4\), \(y=3\). The region is a trapezoid with parallel sides \(1\) and \(3\).
- Find the width. The interval \([0,4]\) has width \(h=4\).
- Apply the trapezoid formula. \(A=\tfrac12(b_1+b_2)h = \tfrac12(1+3)(4)\).
- Compute. \(=\tfrac12(4)(4)=8\) square units.
This matches the definite integral \(\int_0^4(\tfrac12 x+1)\,dx = 8\) exactly — geometry and calculus agree. (It's one of the lab's functions; check it there.)
Try it 1 — Find the area of a rectangle with width \(0.5\) and height \(4\).
Area \(= \text{width}\times\text{height} = 0.5\times 4 = 2\). (This is exactly one Riemann rectangle of width \(\Delta x=0.5\).)
Try it 2 — Find the area of the triangle under \(y=x\) from \(x=0\) to \(x=4\).
The region is a right triangle with base \(4\) and height \(4\) (since \(y=4\) at \(x=4\)). \(A=\tfrac12 bh = \tfrac12(4)(4)=8\). This equals \(\int_0^4 x\,dx = \left[\tfrac{x^2}{2}\right]_0^4 = 8\). ✓
Quick Readiness Check
Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.
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Find an antiderivative of \(f(x)=x^3\).
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Show answer\(\displaystyle\int x^3\,dx = \dfrac{x^4}{4} + C\). (Power \(3\to 4\), divide by \(4\).)
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Why must an indefinite integral include \(+C\)?
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Show answerBecause the derivative of any constant is \(0\), infinitely many functions share the same derivative — they differ only by a constant. The \(+C\) names the whole family.
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Evaluate \(\displaystyle\sum_{i=1}^{3} (2i+1)\).
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Show answerTerms: \(3,5,7\). Sum \(=15\).
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On \([0,3]\), does \(f(x)=x^2\) go below the \(x\)-axis?
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Show answerNo — \(x^2\ge 0\) for every \(x\), so the whole region sits above the axis (all positive area).
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Find the area of a trapezoid with parallel sides \(2\) and \(6\) and width \(4\).
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Show answer\(A=\tfrac12(b_1+b_2)h = \tfrac12(2+6)(4) = \tfrac12(8)(4) = 16\).
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A Riemann rectangle has width \(\Delta x = 1\) and height \(f(2)=4\). What is its area?
Show answer
Show answerArea \(=\) height \(\times\) width \(= 4\times 1 = 4\). (Stack up \(n\) of these and you have a Riemann sum.)
Tools sharpened. When the skills above feel steady, step into the module — the Visual Lab lets you see a Riemann sum converge to the exact integral, right before your eyes.