Unit 08 · Term 2

The Fundamental Theorem of Calculus

The theorem that ties the whole course together: differentiation and integration are inverse operations. It is what lets us compute integrals without ever returning to sums.

By the end you can

State both parts of the FTC.
Explain why the area function's derivative is $f$.
Evaluate definite integrals with antiderivatives.
Differentiate integrals with variable limits.

Why it matters & where it comes from

Why it exists

It would be astonishing if two operations invented for unrelated reasons — slopes and areas — turned out to be exact inverses. They do. That is the Fundamental Theorem, and it is why integration becomes "find an antiderivative" instead of summing forever.

In the real world

Because of it we can solve the differential equations that model nearly everything — orbits, circuits, epidemics (SIR models), and option prices (Black–Scholes). Without it those would be intractable.

Story & open questions

Newton's own teacher, Isaac Barrow, glimpsed the connection geometrically; Newton and Leibniz saw it in full. This is the theorem that turned calculus from a collection of tricks into the central tool of science. The name is not an exaggeration.

The two directions of the theorem

FTC Part 1 says integration creates an antiderivative. FTC Part 2 says an antiderivative evaluates an integral. Keeping those directions separate prevents most mistakes in this unit.

Which part should you use?

If the problem asks for a derivative of $\int_a^x f(t)\,dt$, use Part 1 and the chain rule.

If the problem asks for the numerical value of $\int_a^b f(x)\,dx$, find an antiderivative and use Part 2.

If both limits move, rewrite the integral as a difference of two accumulation functions or apply the upper-limit rule minus the lower-limit rule.

Variable-limit checklist

Upper limit $g(x)$ contributes $f(g(x))g'(x)$.
Lower limit $h(x)$ contributes $-f(h(x))h'(x)$.
The integration variable is a dummy variable; it should not be confused with the outside variable $x$.
If the integrand also contains $x$ outside the dummy variable, the simple FTC rule may not apply directly.

Part 1 — the area function

Theorem FTC, Part 1

If $f$ is continuous on $[a,b]$ and $F(x)=\displaystyle\int_a^x f(t)\,dt$, then $F$ is differentiable on $(a,b)$ and $F'(x)=f(x)$. At the endpoints, the corresponding one-sided derivative has the same value.

Proof idea

$\dfrac{F(x+h)-F(x)}{h}=\dfrac1h\displaystyle\int_x^{x+h}f$ is an average of $f$ over $[x,x+h]$.

As $h\to0$, continuity forces that average to $f(x)$. So $F'(x)=f(x)$. ∎

Part 2 — evaluation

Theorem FTC, Part 2

If $f$ is continuous and $G$ is any antiderivative of $f$ (so $G'=f$), then $$\int_a^b f(x)\,dx=G(b)-G(a).$$

Example

$\displaystyle\int_1^e\frac{\ln x}{x}\,dx$. With $u=\ln x$ an antiderivative is $\tfrac12(\ln x)^2$, so the value is $\tfrac12(1-0)=\tfrac12$.

Variable limits and the chain rule

Rule

$\dfrac{d}{dx}\displaystyle\int_{a}^{g(x)}f(t)\,dt=f\!\big(g(x)\big)\,g'(x).$ With both limits variable, differentiate each and subtract.

Example

$\dfrac{d}{dx}\displaystyle\int_{0}^{x^2}\cos t\,dt=\cos(x^2)\cdot2x.$

Common pitfalls

• In Part 1 the variable is the upper limit; the dummy variable inside is different ($t$, not $x$).
• With a variable upper limit $g(x)$, don't forget the $g'(x)$ factor.
• Part 2 needs $f$ continuous on $[a,b]$ and an antiderivative $G$ with $G'=f$ throughout $(a,b)$.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · FTC part 1

Compute $\dfrac{d}{dx}\displaystyle\int_{x}^{3}e^{-t^2}\,dt$.

Answer

The variable is the lower limit, so it is $-e^{-x^2}$.

Practice 2 · FTC part 2

Evaluate $\displaystyle\int_0^{\pi}\sin x\,dx$.

Answer

$[-\cos x]_0^{\pi}= -(-1)-(-1)=2$.

Practice 3 · Chain rule

Find $F'(x)$ if $F(x)=\displaystyle\int_{1}^{x^3}\frac{dt}{1+t^2}$.

Answer

$\dfrac{1}{1+x^6}\cdot3x^2$.

Practice 4 · Recover f

If $\displaystyle\int_0^x f(t)\,dt=x\,e^{x}$, find $f$.

Answer

Differentiate both sides: $f(x)=e^x+xe^x=(1+x)e^x$.

Practice 5 · Both limits

Find $F'(x)$ for $F(x)=\displaystyle\int_{2x}^{x^2}\frac{dt}{1+t^2}$.

Answer

Differentiate each limit: $F'(x)=\dfrac{2x}{1+x^4}-\dfrac{2}{1+4x^2}.$

Practice 6 · Evaluate

Evaluate $\displaystyle\int_{\pi/4}^{\pi/3}\sec^2x\,dx$.

Answer

$[\tan x]_{\pi/4}^{\pi/3} =\sqrt3-1.$

Practice 7 · Evaluate

Evaluate $\displaystyle\int_0^1\!\left[e^x+e^{-x}-\cos(2x)\right]dx$.

Answer

$(e-1)+(1-e^{-1})-\tfrac{\sin2}{2}=e-e^{-1}-\tfrac{\sin2}{2}.$

Practice 8 · Recover f

Suppose $f$ is continuous and $\displaystyle\int_0^x e^t f(t)\,dt =\frac{\sin x}{x^2+1}$ for all $x$. Find $f(x)$.

Solution

Differentiate: $e^x f(x)=\dfrac{\cos x\,(x^2+1)-2x\sin x}{(x^2+1)^2}$, so $f(x)=e^{-x}\dfrac{\cos x\,(x^2+1)-2x\sin x}{(x^2+1)^2}$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

FTC Part 1. Differentiate accumulation functions and handle variable upper or lower limits.
FTC Part 2. Evaluate definite integrals using antiderivatives under the right continuity hypotheses.
Net change. Interpret integrals of rates as total change.
Substitution connection. See substitution for definite integrals as a consequence of the chain rule and FTC.

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The spine is: accumulation creates antiderivatives, antiderivatives evaluate accumulation, and the two halves close the loop between differentiation and integration.

Study routine. For every FTC problem, label which part of the FTC is being used; many errors come from applying Part 2 when the task needs Part 1 plus chain rule.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Variable endpoint

Differentiate $F(x)=\int_2^{x^3}\sqrt{1+t^4}\,dt$.

Guide

FTC Part 1 plus chain rule gives $F\prime(x)=3x^2\sqrt{1+x^{12}}$.

Extension 2 · Lower endpoint

Differentiate $G(x)=\int_x^1 e^{-t^2}\,dt$.

Guide

Rewrite as $-\int_1^x e^{-t^2}\,dt$, so $G\prime(x)=-e^{-x^2}$.

Extension 3 · Net change

If $v(t)=3t^2$ meters/sec, find displacement from $t=1$ to $t=3$.

Guide

Compute $\int_1^3 3t^2dt=[t^3]_1^3=26$ meters.

Extension 4 · Hypothesis check

Why does FTC Part 1 require continuity in the usual statement?

Guide

Continuity ensures the average value of the integrand over shrinking intervals approaches the point value; without it, the accumulation function may fail to have derivative equal to the integrand at discontinuities.

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