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Fundamental Theorem of Calculus cheat sheet - grade 11-12

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Calculus Grade 11-12

Fundamental Theorem of Calculus Cheat Sheet

A printable reference covering FTC Part 1, FTC Part 2, accumulation functions, net change, and average value for grades 11-12.

TopicsIntegralsDerivativesFunctionsCalculusStudy Guide
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The Fundamental Theorem of Calculus connects derivatives and definite integrals, showing that accumulation and rate of change are inverse ideas. This cheat sheet helps students recognize when to differentiate an integral, evaluate a definite integral using an antiderivative, or interpret accumulated change. It is especially useful for AP Calculus and precalculus-to-calculus review because it links graphs, formulas, and units.

The main ideas are FTC Part 1, FTC Part 2, net change, and accumulation functions. FTC Part 1 says that if F(x)=∫axf(t) dtF(x)=\int_a^x f(t)\,dt, then F′(x)=f(x)F'(x)=f(x) when ff is continuous. FTC Part 2 says that ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a) when F′(x)=f(x)F'(x)=f(x).

These formulas explain why integrals measure accumulated change and derivatives measure instantaneous change.

Key Facts

  • FTC Part 1 states that if F(x)=∫axf(t) dtF(x)=\int_a^x f(t)\,dt and ff is continuous, then F′(x)=f(x)F'(x)=f(x).
  • FTC Part 2 states that if F′(x)=f(x)F'(x)=f(x), then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a).
  • For a variable upper limit, ddx∫ag(x)f(t) dt=f(g(x))g′(x)\frac{d}{dx}\int_a^{g(x)} f(t)\,dt=f(g(x))g'(x) by the chain rule.
  • For a variable lower limit, ddx∫g(x)af(t) dt=−f(g(x))g′(x)\frac{d}{dx}\int_{g(x)}^a f(t)\,dt=-f(g(x))g'(x).
  • Net change is calculated by ∫abr(t) dt=Q(b)−Q(a)\int_a^b r(t)\,dt=Q(b)-Q(a) when r(t)=Q′(t)r(t)=Q'(t).
  • Total change uses accumulated distance or amount and may require ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt instead of ∫abv(t) dt\int_a^b v(t)\,dt.
  • The average value of a continuous function on [a,b][a,b] is favg=1b−a∫abf(x) dxf_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.
  • A definite integral can be negative when the graph of f(x)f(x) lies below the xx-axis, because signed area is counted.

Vocabulary

Definite integral
A definite integral ∫abf(x) dx\int_a^b f(x)\,dx represents the signed accumulation of f(x)f(x) from x=ax=a to x=bx=b.
Antiderivative
An antiderivative of f(x)f(x) is a function F(x)F(x) such that F′(x)=f(x)F'(x)=f(x).
Accumulation function
An accumulation function has the form A(x)=∫axf(t) dtA(x)=\int_a^x f(t)\,dt and gives the accumulated signed area up to xx.
Net change
Net change is the final amount minus the initial amount, written as Q(b)−Q(a)=∫abQ′(t) dtQ(b)-Q(a)=\int_a^b Q'(t)\,dt.
Signed area
Signed area counts regions above the horizontal axis as positive and regions below the horizontal axis as negative.
Average value
The average value of f(x)f(x) on [a,b][a,b] is 1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx.

Common Mistakes to Avoid

  • Forgetting the chain rule with variable limits is wrong because ddx∫ag(x)f(t) dt\frac{d}{dx}\int_a^{g(x)} f(t)\,dt equals f(g(x))g′(x)f(g(x))g'(x), not just f(g(x))f(g(x)).
  • Treating all definite integrals as positive area is wrong because ∫abf(x) dx\int_a^b f(x)\,dx measures signed area, so parts below the axis subtract.
  • Using the integrand instead of an antiderivative in FTC Part 2 is wrong because ∫abf(x) dx\int_a^b f(x)\,dx requires F(b)−F(a)F(b)-F(a) where F′(x)=f(x)F'(x)=f(x).
  • Confusing net change with total distance is wrong because ∫abv(t) dt\int_a^b v(t)\,dt gives displacement, while ∫ab∣v(t)∣ dt\int_a^b |v(t)|\,dt gives total distance.
  • Dropping the negative sign for a variable lower limit is wrong because ddx∫g(x)af(t) dt=−f(g(x))g′(x)\frac{d}{dx}\int_{g(x)}^a f(t)\,dt=-f(g(x))g'(x).

Practice Questions

  1. 1 Evaluate ∫132x dx\int_1^3 2x\,dx using the Fundamental Theorem of Calculus.
  2. 2 If A(x)=∫2x(t2+1) dtA(x)=\int_2^x (t^2+1)\,dt, find A′(x)A'(x) and A′(3)A'(3).
  3. 3 Find ddx∫0x21+t3 dt\frac{d}{dx}\int_0^{x^2} \sqrt{1+t^3}\,dt.
  4. 4 A velocity graph is sometimes above and sometimes below the time axis on [0,6][0,6]. Explain why ∫06v(t) dt\int_0^6 v(t)\,dt and ∫06∣v(t)∣ dt\int_0^6 |v(t)|\,dt can represent different physical quantities.

Understanding Fundamental Theorem of Calculus

A useful way to understand an accumulation function is to imagine moving its upper boundary a tiny amount to the right. The added piece of accumulated area is a very thin rectangle. Its height is close to the value of the original function, while its width is the tiny change in the input.

Dividing the added area by that tiny width leaves the height. This is the reason the derivative of an accumulation function gives the current value of the rate. Continuity matters because a graph with a jump can make the added sliver behave less predictably at that point.

Variable limits require careful reading because the input can affect more than one part of the expression. When the upper boundary is a function, first evaluate the integrand at that boundary. Then account for how fast the boundary itself moves.

If the boundary moves twice as fast, it sweeps out accumulated area twice as fast. A lower boundary has the opposite direction. Increasing it removes area from the interval rather than adding area.

Students often lose the negative sign because they picture only the size of the region. Sketching the interval before differentiating makes the direction clear.

Net change has a strong connection to real measurements. A velocity graph records signed velocity, so its integral gives displacement. Positive motion and negative motion can cancel.

A person who walks thirty meters east then thirty meters west has zero displacement, even though they traveled sixty meters. For total distance, each piece of velocity must contribute a positive amount. This means finding where velocity changes sign, splitting the time interval there, and treating each travel segment by its size.

The same distinction appears with profit and loss, water entering and leaving a tank, electric charge, and population changes. Units provide an important check. A rate in liters per minute accumulated over minutes produces liters.

Average value describes a constant height that would produce the same accumulated amount across an interval. On a graph, it is the height of a rectangle whose area matches the signed area under the curve. It is not usually found by averaging just the endpoint values.

That shortcut works only in special cases, such as a linear function over an interval. For continuous functions, there is at least one point where the actual function reaches its average value. When solving problems, keep track of whether an integral represents signed area, total amount, or a change in a quantity.

Use an antiderivative only after checking the bounds and the variable. Substitute the upper endpoint first, then subtract the lower endpoint. This order prevents many common sign errors.