Differentiation: rates of change (instantaneous)

Integration: accumulation (net change / area)

Differentiation and integration are inverse processes (under appropriate conditions)

\(\int_a^b f(x)\,dx\): net accumulation of \(f\) from \(a\) to \(b\)

Interpreted as signed area when \(f\ge 0\) (or net area generally)

\(F'(x)=f(x)\) means \(F\) is an antiderivative of \(f\)

\(\int f(x)\,dx = F(x)+C\)

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

Measures accumulated value of \(f\) from \(a\) to \(x\)

FTC Part 1: \(f\) continuous on \([a,b]\) (or weaker: integrable + mild conditions)

FTC Part 2: \(f\) continuous and \(F\) differentiable with \(F'=f\)

If \(f\) is continuous on \([a,b]\) and \(A(x)=\int_a^x f(t)\,dt\), then

The instantaneous rate of change of accumulated area equals the integrand value

Accumulation function is an antiderivative of \(f\)

Variable upper limit with scaling/shift

Variable lower limit

Both limits variable

Integrand depends on \(x\) and \(t\) (Leibniz rule, common extension)

Small increase \(\Delta x\) adds approximately a rectangle of height \(f(x)\) and width \(\Delta x\)

So \(\Delta A \approx f(x)\Delta x\) ⇒ \(A'(x)=f(x)\)

If \(f\) is continuous on \([a,b]\) and \(F\) is any antiderivative of \(f\), then

Net accumulation equals net change in an antiderivative over the interval

Converts area/accumulation problems into antiderivative evaluation

Find \(F\) such that \(F'=f\)

Compute \(F(b)-F(a)\)

Common shorthand: \(\int_a^b f(x)\,dx = \big[F(x)\big]_a^b\)

Any two antiderivatives differ by a constant

The constant cancels in \(F(b)-F(a)\), making definite integrals well-defined

Shows \(\int_a^x f(t)\,dt\) produces an antiderivative of \(f\)

Uses any antiderivative to compute the definite integral

Accumulation function \(A(x)\) is an antiderivative ⇒ \(\int_a^b f = A(b)-A(a)\)

Since \(A(a)=0\), \(\int_a^b f = A(b)\)

If \(F'(x)=f(x)\), then

Derivative gives instantaneous rate

Integral gives total accumulation of that rate over time/space

Positive where \(f>0\), negative where \(f<0\)

Integral measures net effect, not necessarily geometric area

Area under curve (when \(f\ge 0\))

Net area / displacement (when sign changes)

Velocity \(v(t)\) ⇒ displacement \(\int v(t)dt\)

Flow rate ⇒ total volume

Marginal cost ⇒ total cost change

Given \(A(x)=\int_a^x f(t)\,dt\), evaluate \(A'(x)\), find increasing/decreasing behavior

Sensitivity of accumulated quantities to changing bounds

\(A(x)=\int_0^x \cos t\,dt \Rightarrow A'(x)=\cos x\)

\(\frac{d}{dx}\int_1^{x^2} \sqrt{1+t^4}\,dt = \sqrt{1+x^8}\cdot 2x\)

\(\int_0^{\pi} \sin x\,dx = [-\cos x]_0^{\pi} = 2\)

\(\int_1^4 \frac{1}{\sqrt{x}}\,dx = [2\sqrt{x}]_1^4 = 2\)

Indefinite: family \(F(x)+C\)

Definite: number \(F(b)-F(a)\)

Upper limit \(g(x)\) introduces factor \(g'(x)\)

\(\int_a^b f\) is net signed accumulation

Total geometric area often requires \(\int_a^b |f(x)|dx\)

Continuity guarantees clean statements

For discontinuities, FTC can hold in weaker forms but needs careful hypotheses

\(\frac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)\)

\(\int_a^b f(x)\,dx=F(b)-F(a)\) when \(F'=f\)

Integration accumulates; differentiation recovers the original rate (inverse relationship under suitable conditions)