Part IV · Series, Exponentials, and Integral Calculus

Chapter 12 — Definite Integrals and the Fundamental Theorem

Why This Matters

Derivatives describe local rate. Integration reverses the question: if a rate is known, how much has accumulated? Current is rate of charge flow, so accumulated charge is an integral of current. Area is another familiar accumulation model.

The Fundamental Theorem connects rate of change to accumulation. Differentiation computes instantaneous rates; integration accumulates those rates over an interval. They are exact inverses.

Fundamental Theorem of Calculus (Part 1 and 2)

Starting point. Let $$f$$ be continuous on $$[a, b]$$ , and define the accumulator $g(x) = \int_a^x f(t) dt$ .

\[g'(x) = \lim_{h\to0} \frac{g(x+h) - g(x)}{h} = \lim_{h\to0} \frac{1}{h} \int_x^{x+h} f(t) dt\]
Write the derivative of the accumulation function by definition.reversible
\[\lim_{h\to0} \frac{1}{h} \left[ f(c) \cdot h \right] = \lim_{h\to0} f(c) = f(x)\]
By the Mean Value Theorem for Integrals, $\int_x^{x+h} f(t) dt = f(c) h$ for some $c \in [x, x+h]$ . As $h \to 0$ , $c \to x$ .reversible
\[\int_a^b f(t) dt = F(b) - F(a)\]
Since $$g(x)$$ is an antiderivative of $$f(x)$$ , any antiderivative $$F$$ satisfies $\int_a^b f(t)dt = F(b) - F(a)$ .reversible

Conclusion. FTC Part 1: $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ . FTC Part 2: $\int_a^b f(x) dx = F(b) - F(a)$ .

12.1 Riemann sums and signed area

For a function $$f$$ on $$[a,b]$$ , split the interval into narrow pieces of width $\Delta x$ . A rectangle of height $$f(x_k^*)$$ has signed area $f(x_k^*)\Delta x$ . Adding the rectangles and taking the limit defines

\int_a^b f(x)\,dx.

Area above the axis contributes positively and area below contributes negatively. The integral is therefore signed accumulation, not automatically total geometric area.

Geometric picture. A definite integral is built from ordinary rectangle areas. Narrower rectangles reduce the mismatch between the curve and each rectangle; portions below the axis contribute negative signed area.

\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_k^*)\Delta x

Left and right rectangles approximating area under y equals x squared.

As the number of rectangles increases, the Riemann sum approaches the exact area 1/3 under y = x^2 on [0, 1].

12.2 Antiderivatives and the FTC

An antiderivative of $$f$$ is a function $$F$$ with $$F'=f$$ . Antiderivatives differ by a constant because constants differentiate to zero.

The theorem makes exact integration practical. It also says that the accumulation function $A(x)=\int_a^x f(t)\,dt$ has derivative $$A'(x)=f(x)$$ when $$f$$ is continuous.

Example — accumulated charge

If $$i(t)=(2t+1)$$ mA, with $$t$$ measured in milliseconds and $0\le t\le3$ , then the charge is

q=\int_0^3(2t+1)\ \mathrm{mA\,ms} =\left[t^2+t\right]_0^3\ \mathrm{\mu C} =12\ \mathrm{\mu C}.

The unit conversion is built into $\mathrm{mA\,ms}=10^{-6}\ \mathrm C$ . Writing the time unit explicitly also prevents a common factor-of- $$1000$$ error.

Accumulation produces rate. Let A(x) be the signed area accumulated from a fixed starting point to x. Extending the endpoint by a tiny amount adds a very thin strip whose area is approximately its height f(x) times its width. Dividing by that width and taking the limit gives A'(x)=f(x).

A(x)=\int_a^x f(t)\,dt\quad\Longrightarrow\quad A'(x)=f(x)

Accumulated area under a curve, approximated by an adjustable number of rectangles.

As the rectangle count increases, the Riemann sum converges to the accumulated area 1/3 under y = x^2 on [0, 1].

12.3 Basic antiderivatives

For $a\ne-1$ ,

\int x^a\,dx=\frac{x^{a+1}}{a+1}+C.

The excluded exponent has its own rule:

\int\frac1x\,dx=\ln|x|+C

on intervals not crossing zero. Also,

\int e^{ax}\,dx=\frac1a e^{ax}+C\quad(a\ne0).

Antidifferentiation is the reverse of differentiation. If $$F'(x) = f(x)$$ , then $\int f(x) dx = F(x) + C$ . The constant $$+C$$ represents a family of vertically shifted parallel curves all sharing the same slope at every $$x$$ .

Why the constant of integration is necessary

Starting point. Let $$F(x)$$ and $$G(x)$$ be two antiderivatives of $$f(x)$$ on an interval.

\[\frac{d}{dx}[F(x) - G(x)] = F'(x) - G'(x) = f(x) - f(x) = 0\]
The derivative of their difference is identically zero.reversible
\[F(x) - G(x) = C\]
By the Mean Value Theorem, a function whose derivative is everywhere zero on an interval must be a constant.reversible
\[F(x) = G(x) + C\]
Any two antiderivatives differ by at most an additive constant.reversible

Conclusion. The indefinite integral $\int f(x)dx = F(x) + C$ captures all possible antiderivatives. Omitting $$+C$$ loses solutions to differential equations.

Power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for all $n \neq -1$ .
Reciprocal rule: $\int \frac{1}{x} dx = \ln|x| + C$ .
Exponential rule: $\int e^{kx} dx = \frac{1}{k} e^{kx} + C$ .
Always add $$+C$$ to indefinite integrals.

12.4 Substitution

Substitution reverses the chain rule. If $$u=g(x)$$ and $du=g'(x)\,dx$ , then

\int f(g(x))g'(x)\,dx=\int f(u)\,du.

For a definite integral, either change the limits to $$u$$ -limits or return to $$x$$ before substituting the original limits.

Example — substitution with changed bounds

\int_0^1 2x e^{x^2}\,dx.

Let $$u=x^2$$ , so $du=2x\,dx$ . The bounds become $$u=0$$ and $$u=1$$ :

\int_0^1e^u\,du=e-1.

Substitution names the changing inner quantity. When an inner expression changes by a known derivative factor, replacing it by a new variable turns a composite accumulation into a simpler one-variable accumulation. For a definite integral, the limits must change too because they describe values of the new variable.

Choose the repeated inner expression as u.
Write du with its differential factor and account for any constant multiplier.
For definite bounds, convert each original endpoint to its u-value before evaluating.

12.5 Building a definite integral from rectangles

The symbol $\int_a^b f(x)\,dx$ is compact, but it represents a very concrete limiting measurement. Suppose $$f(x)=x^2$$ on $$[0,1]$$ . Divide the interval into $$n$$ equal pieces. Each has width $\Delta x=1/n$ . If right endpoints are used, the $$k$$ th height is $$(k/n)^2$$ , so the right-rectangle total is

S_n=\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac1n =\frac1{n^3}\sum_{k=1}^n k^2.

The finite sum formula $\sum_{k=1}^n k^2=n(n+1)(2n+1)/6$ gives

S_n=\frac{n(n+1)(2n+1)}{6n^3} =\frac{2n^3+3n^2+n}{6n^3}.

As $n\to\infty$ , the lower-order terms disappear and $S_n\to1/3$ . Thus

\int_0^1x^2\,dx=\frac13.

This calculation is not a replacement for the Fundamental Theorem of Calculus; it explains what that theorem evaluates efficiently. A graph should be imagined here: the curve $$y=x^2$$ rises from $$(0,0)$$ to $$(1,1)$$ , while the narrow rectangles leave ever smaller gaps above or below the curve.

[Figure: For

\(y=x^2\)

\([0,1]\)

, show left and right Riemann rectangles for

$$n=4$$ , then a second panel with many thin rectangles. Label $\Delta x$ and a sample height $$f(x_k^*)$$ .]

Example — signed accumulation is not total area

Evaluate

\int_{-2}^{2}x\,dx.

An antiderivative is $$x^2/2$$ , so

\left[\frac{x^2}{2}\right]_{-2}^{2}=2-2=0.

The result follows also from symmetry: the triangular region below the axis on $$[-2,0]$$ has the same magnitude as the triangular region above it on $$[0,2]$$ . If the question asks for geometric area instead, split at the zero and reverse the sign on the negative part:

\int_{-2}^{0}(-x)\,dx+\int_0^2x\,dx=2+2=4.

Always decide which quantity the model asks for. Net charge can be signed; the total distance travelled cannot be obtained from velocity by allowing opposite directions to cancel.

Riemann sums construct the exact integral. The limit of Riemann sums as $\Delta x \to 0$ ( $n \to \infty$ ) defines the integral regardless of the sample point choice (left, right, or midpoint).

Definite integral as Riemann limit

Starting point. Divide $$[a, b]$$ into $$n$$ equal subintervals of width $\Delta x = \frac{b-a}{n}$ with grid points $x_k = a + k \Delta x$ .

\[R_n = \sum_{k=1}^n f(x_k) \Delta x\]
Right Riemann sum: sample heights at right endpoints $$x_k$$ .reversible
\[L_n = \sum_{k=0}^{n-1} f(x_k) \Delta x\]
Left Riemann sum: sample heights at left endpoints $$x_k$$ .reversible
\[\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{k=1}^n f(x_k^*) \Delta x\]
As $n \to \infty$ and $\Delta x \to 0$ , all Riemann sums converge to the same unique real number for any continuous $$f$$ .reversible

Conclusion. $\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{k=1}^n f(x_k^*) \Delta x$ . The integral represents the net signed area bounded by $$y = f(x)$$ and the $$x$$ -axis.

12.6 Why the Fundamental Theorem works

There are two complementary parts to the theorem. The first says that an accumulation function has the original rate as its derivative. Define

A(x)=\int_a^x f(t)\,dt.

The letter $$t$$ is a dummy variable: changing it to $$u$$ does not change the meaning. For a small change $$h$$ ,

\frac{A(x+h)-A(x)}{h} =\frac1h\int_x^{x+h}f(t)\,dt.

The right-hand side is the average value of $$f$$ over a short interval. When $$f$$ is continuous, every value in that short interval becomes close to $$f(x)$$ as $h\to0$ . Therefore $$A'(x)=f(x)$$ . This is the accumulation analogue of the secant line becoming a tangent line.

The second part says that if $$F'=f$$ , then the total accumulation from $$a$$ to $$b$$ is $$F(b)-F(a)$$ . Since $$A'=F'$$ , the functions $$A$$ and $$F$$ differ by a constant. Because $$A(a)=0$$ , that constant is $$-F(a)$$ , and hence

\[A(b)=F(b)-F(a).\]

The continuity hypothesis gives a dependable ordinary Riemann integral and makes the average-value argument valid. More advanced mathematics extends the theorem further, but this course uses the stated continuous-function setting.

Example — an accumulation function before evaluating it

Let

A(x)=\int_1^x\left(3t^2-4\right)\,dt.

Without first performing any integration, the Fundamental Theorem gives

\[A'(x)=3x^2-4.\]

To find the function itself, use $$F(t)=t^3-4t$$ :

\[A(x)=(x^3-4x)-(1-4)=x^3-4x+3.\]

Checking $$A(1)=0$$ is essential. An accumulation from a point to itself must be zero; this catches a lost lower-limit term.

Why differentiation and integration undo each other. The Fundamental Theorem links the local slope of an area function directly to the height of the integrand.

Proof of FTC Part 1

Assumptions

$$f$$ is continuous on $$[a, b]$$ .
Define $A(x) = \int_a^x f(t) dt$ .

$A(x+h) - A(x) = \int_x^{x+h} f(t) dt$ .By interval additivity of definite integrals.
$\int_x^{x+h} f(t) dt = f(c) h$ for some $c \in [x, x+h]$ .By the Mean Value Theorem for Integrals (since $$f$$ is continuous).
$A'(x) = \lim_{h\to0} \frac{A(x+h) - A(x)}{h} = \lim_{h\to0} f(c) = f(x)$ .As $h \to 0$ , the squeeze $x \le c \le x+h$ forces $c \to x$ , and by continuity $f(c) \to f(x)$ .

Conclusion. $\frac{d}{dx} \int_a^x f(t) dt = f(x)$ . The rate of area accumulation equals the instantaneous curve height.

Scope. Requires $$f$$ to be continuous at $$x$$ .

12.7 Substitution as a change of scale

The chain rule says

\frac d{dx}F(g(x))=F'(g(x))g'(x).

Substitution reads this equation backwards. The inner expression $$g(x)$$ is a new coordinate, and $g'(x)\,dx$ tells us how a small change in $$x$$ changes that coordinate. It is not enough for an integrand to contain something that looks like an inner expression: the accompanying differential must match, possibly after extracting a constant.

Example — recognise and repair the missing factor

Find

\int x\cos(x^2)\,dx.

The inner expression is $$x^2$$ and its derivative is $$2x$$ . Write the available factor as $x\,dx=\tfrac12(2x\,dx)$ , then set $$u=x^2$$ :

\int x\cos(x^2)\,dx =\frac12\int\cos u\,du =\frac12\sin u+C =\frac12\sin(x^2)+C.

Differentiation confirms the result: the outer factor $$1/2$$ and chain-rule factor $$2x$$ combine to recover $x\cos(x^2)$ .

Example — a definite integral with two safe routes

Evaluate

\int_1^2\frac{2x}{x^2+3}\,dx.

With $$u=x^2+3$$ , $du=2x\,dx$ . Changing bounds gives $$u=4$$ at $$x=1$$ and $$u=7$$ at $$x=2$$ , so

\int_4^7\frac1u\,du=[\ln u]_4^7=\ln\frac74.

Equally, one may return to $$x$$ first and calculate $[\ln(x^2+3)]_1^2=\ln7-\ln4$ . The unsafe hybrid is to use $$u$$ but retain the old bounds $$1$$ and $$2$$ ; those numbers name $$x$$ -positions, not $$u$$ -values.

$$u$$ -substitution scales the coordinate axis. Substituting $$u = g(x)$$ transforms the integration variable; the differential relation $$du = g'(x) dx$$ accounts for the local stretching or compression of coordinates.

12.8 A reliable integral-solving routine

Before calculating, identify the quantity being accumulated and attach units. Then inspect the integrand: a sum invites term-by-term integration; a composite expression with its inner derivative invites substitution; a product may need integration by parts in Chapter 13. For a definite integral, sketch signs, zeros, and any discontinuities. Finally differentiate an indefinite answer or check an exact definite result by a size estimate.

Engineering Example — charge and average current

A current is $$i(t)=4-0.5t$$ mA for $0\le t\le6$ ms. The charge delivered is

Q=\int_0^6(4-0.5t)\,dt\ \mathrm{mA\,ms} =\left[4t-0.25t^2\right]_0^6\ \mu\mathrm C =15\ \mu\mathrm C.

The average current over the interval is total charge divided by elapsed time:

i_{\rm avg}=\frac{Q}{6\ \mathrm{ms}}=2.5\ \mathrm{mA}.

This is also the average height of the graph. Since current falls linearly from $$4$$ mA to $$1$$ mA, the same answer is the mean of the endpoint values. That geometric check is available only because the graph is a straight line.

$$u$$ -substitution is the chain rule in reverse. If an integral contains an inner function $$g(x)$$ alongside a multiple of its derivative $$g'(x)$$ , substitution converts it into a simpler integral in terms of $$u$$ .

Substitution formula

Starting point. By the chain rule, $\frac{d}{dx}[F(g(x))] = F'(g(x)) g'(x) = f(g(x)) g'(x)$ .

\[\int f(g(x)) g'(x) dx = F(g(x)) + C\]
Integrate both sides with respect to $$x$$ .reversible
\[\text{Let } u = g(x) \implies du = g'(x) dx\]
Substitute variable $$u$$ and differential $$du$$ .reversible
\[\int f(u) du = F(u) + C = F(g(x)) + C\]
Evaluate the simpler integral in $$u$$ , then substitute back $$u = g(x)$$ .reversible

Conclusion. $\int f(g(x)) g'(x) dx = \int f(u) du$ . For definite integrals, transform the integration limits: $\int_a^b f(g(x))g'(x)dx = \int_{g(a)}^{g(b)} f(u)du$ .

Identify the 'inner' function $$u = g(x)$$ whose derivative $$g'(x)$$ appears as a multiplying factor.
Calculate the differential: $du = g'(x) dx \implies dx = \frac{du}{g'(x)}$ .
Substitute $$u$$ and $$du$$ , ensuring all instances of $$x$$ cancel completely.
For definite integrals, update limits: $$u_1 = g(a)$$ and $$u_2 = g(b)$$ (no need to back-substitute).
For indefinite integrals, evaluate in $$u$$ , then back-substitute $$u = g(x)$$ .

Exercises

Interpret $\int_0^4 v(t)\,dt$ when $$v$$ is a velocity.
Evaluate $\int_0^2(3x^2-4x+1)\,dx$ .
Find $\int(5x-1)^7\,dx$ by substitution.
Explain why $\int_{-1}^1x\,dx=0$ does not mean the geometric area is zero.
A current is $i(t)=I_0e^{-t/\tau}$ . Find total charge delivered from $$0$$ to $$T$$ .
Use a right Riemann sum to derive $\int_0^1x\,dx=1/2$ .
Let $A(x)=\int_0^x\sin t\,dt$ . Find $$A'(x)$$ and $A(\pi)$ .
Evaluate $\int_0^1 3x^2/(1+x^3)\,dx$ , showing the changed limits.
A velocity is $$v(t)=t^2-4t+3$$ for $0\le t\le3$ . Find displacement and total distance; explain why they differ.