MAT158H1/Unit 10

Unit 10 · Analysis 1A

The Fundamental Theorem of Calculus

Differentiation and integration are inverse operations — proved here from the Riemann definition, using the EVT for Part 1 and the Mean Value Theorem for Part 2. Everything built since Unit 2 comes together on this page.

By the end you can

• State and prove both parts of the FTC.
• Explain which hypotheses each part actually needs.
• Evaluate integrals via antiderivatives.
• Differentiate integrals with variable limits.

Why it matters & where it comes from

Why it exists

Two operations invented for unrelated reasons — slopes and areas — turn out to be exact inverses. That is what lets us integrate by finding antiderivatives instead of summing forever.

In the real world

It is what makes differential equations solvable — the models behind orbits, circuits, epidemics, and the Black–Scholes option price all rely on it.

Story & open questions

Newton's teacher Isaac Barrow glimpsed the link geometrically; Newton and Leibniz saw it in full. This is the result that turned calculus from a bag of tricks into the central tool of science.

Part 1 — the area function has derivative $f$

Theorem FTC, Part 1

Let $f$ be integrable on $[a,b]$ and $F(x)=\displaystyle\int_a^x f$. If $f$ is continuous at $x$, then $F$ is differentiable at $x$ and $F'(x)=f(x)$.

Proof (for $h>0$; $h<0$ is symmetric)

Additivity gives $F(x+h)-F(x)=\displaystyle\int_x^{x+h}f$. With $m_h=\inf f$ and $M_h=\sup f$ on $[x,x+h]$, monotonicity yields $m_h\,h\le F(x+h)-F(x)\le M_h\,h$, so $m_h\le\dfrac{F(x+h)-F(x)}{h}\le M_h$.

Continuity of $f$ at $x$ forces $m_h\to f(x)$ and $M_h\to f(x)$ as $h\to0$: the difference quotient is squeezed to $f(x)$. ∎

What if $f$ is merely integrable? $F$ is still continuous — in fact Lipschitz: if $|f|\le K$ then $|F(x+h)-F(x)|=\left|\int_x^{x+h}f\right|\le K|h|$. Integration always smooths; differentiating the result needs continuity of $f$ at the point. At a jump of $f$, $F$ has a corner.

Part 2 — evaluation by antiderivatives

Theorem FTC, Part 2

If $f$ is integrable on $[a,b]$ and $G$ is an antiderivative of $f$ on $[a,b]$ (meaning $G'=f$ there), then $$\int_a^b f=G(b)-G(a).$$

Proof — the MVT does all the work

Take any partition $a=x_0<\dots<x_n=b$. On each piece the MVT gives $t_i\in(x_{i-1},x_i)$ with $G(x_i)-G(x_{i-1})=G'(t_i)\,\Delta x_i=f(t_i)\,\Delta x_i$.

Summing telescopes the left side: $G(b)-G(a)=\sum_i f(t_i)\,\Delta x_i$. Since $m_i\le f(t_i)\le M_i$, this lies between $L(f,P)$ and $U(f,P)$ — for every partition.

Integrability says exactly one number sits between all lower and upper sums, namely $\int_a^b f$. So $G(b)-G(a)=\int_a^b f$. ∎

Example

$\displaystyle\int_1^e\frac{\ln x}{x}\,dx$: an antiderivative of $\tfrac{\ln x}{x}$ is $\tfrac12(\ln x)^2$ (check by the chain rule), so the value is $\tfrac12(1^2-0^2)=\tfrac12$.

Variable limits

Rule

For continuous $f$ and differentiable $g$, $\dfrac{d}{dx}\displaystyle\int_a^{g(x)}f(t)\,dt=f\big(g(x)\big)\,g'(x)$ — Part 1 composed with the chain rule. With both limits variable, differentiate each and subtract.

Example

$\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt=\cos(x^2)\cdot2x$ — and you can confirm it by evaluating the integral to $\sin(x^2)$ first.

Common pitfalls

• In Part 1 the variable is the upper limit; the dummy variable inside is a different letter.
• With a variable upper limit $g(x)$, keep the $g'(x)$ factor.
• Part 2 needs $G'=f$ on the whole interval — an antiderivative that misses one point can change the game (corners in $F$ at jumps of $f$).
• Part 1 at a discontinuity of $f$ can fail: $F$ need not be differentiable there.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · Variable limit

Compute $\dfrac{d}{dx}\displaystyle\int_0^{x^2}\cos t\,dt$.

Answer

$\cos(x^2)\cdot2x$, by Part 1 and the chain rule.

Practice 2 · Evaluate

Evaluate $\displaystyle\int_1^e\dfrac{\ln x}{x}\,dx$.

Answer

Antiderivative $\tfrac12(\ln x)^2$; the value is $\tfrac12$.

Practice 3 · Recover $f$

If $f$ is continuous and $\displaystyle\int_0^x f=x e^x$ for all $x$, find $f$.

Answer

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

Practice 4 · Both limits

Find $F'(x)$ for $F(x)=\displaystyle\int_x^{2x}e^{-t^2}\,dt$.

Solution

Split at $0$ (or any point) and apply the rule to each limit: $F'(x)=2e^{-4x^2}-e^{-x^2}$.

Practice 5 · A functional equation

Suppose $f$ is continuous and $\displaystyle\int_0^x f=\int_x^1 f$ for every $x\in[0,1]$. Prove $f\equiv0$.

Solution

Both sides are differentiable in $x$ (Part 1): the left has derivative $f(x)$, the right $-f(x)$. So $f(x)=-f(x)$, forcing $f(x)=0$ for all $x$.

Practice 6 · Evaluate

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

Answer

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

Practice 7 · Chain both ways

Compute $\dfrac{d}{dx}\displaystyle\int_{\sin x}^{1}\sqrt{1+t^2}\,dt$.

Solution

The variable is the lower limit: the derivative is $-\sqrt{1+\sin^2x}\,\cos x$.

Practice 8 · Lipschitz area function

Let $f$ be continuous on $[a,b]$ and $F(x)=\int_a^x f$. Prove there is $M$ with $|F(x)-F(y)|\le M|x-y|$ for all $x,y\in[a,b]$.

Solution

By the EVT, $|f|\le M$ for some $M$. Then $|F(x)-F(y)|=\left|\int_y^x f\right|\le\int_{[y,x]}|f|\le M|x-y|$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

• FTC Part 1. Prove and use differentiability of accumulation functions.
• FTC Part 2. Evaluate integrals using antiderivatives and prove the net-change formula.
• Substitution. Connect substitution to FTC and chain rule.
• Regularity. Track continuity hypotheses for the integrand and differentiability hypotheses for antiderivatives.

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The FTC spine is: continuity gives differentiable accumulation; any antiderivative differs by a constant; definite integrals become endpoint differences.

Study routine. In proof questions, separate the construction of an antiderivative from the uniqueness-up-to-constant argument.

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 · FTC derivative

Differentiate $F(x)=\int_a^x f(t)dt$ when $f$ is continuous.

Guide

FTC Part 1 gives $F\prime(x)=f(x)$.

Extension 2 · Endpoint difference

If $G\prime=f$, prove $\int_a^b f=G(b)-G(a)$.

Guide

By FTC Part 1, $F(x)=\int_a^x f$ is an antiderivative. Then $F-G$ is constant by MVT consequences, and evaluating at $a,b$ gives the formula.

Extension 3 · Chain endpoint

Differentiate $\int_0^{x^2}\cos(t^3)dt$.

Guide

FTC Part 1 plus chain rule gives $2x\cos(x^6)$.

Extension 4 · Substitution

Explain substitution for definite integrals as an FTC result.

Guide

If $u=g(x)$, then an antiderivative of $f(g(x))g\prime(x)$ is $F(g(x))$ where $F\prime=f$; endpoints become $g(a)$ and $g(b)$.

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