Supplement: Rules of Integration and Method Choice

Formula reference. Use $+C$ for indefinite integrals; do not add it to an evaluated definite integral.

Integration rules

$\int[af(x)+bg(x)]\,dx=a\int f(x)\,dx+b\int g(x)\,dx,\qquad \int x^n\,dx=\frac{x^{n+1}}{n+1}+C\quad(n\ne-1).$

$\int\frac1x\,dx=\ln|x|+C,\qquad \int f(g(x))g'(x)\,dx=\int f(u)\,du\quad(u=g(x)).$

$\int_a^bf(x)\,dx=F(b)-F(a)\quad\text{when }F'=f,\qquad Q(b)=Q(a)+\int_a^bQ'(t)\,dt.$

For total distance use $\int_a^b|v(t)|\,dt$; for displacement use $\int_a^bv(t)\,dt$.

Integration accumulates contributions, and an antiderivative reverses differentiation. Those two ideas meet in the Fundamental Theorem of Calculus, but they answer different questions. An indefinite integral asks for a family of functions whose derivative is the integrand. A definite integral asks for total signed accumulation over an interval. Keeping the question visible prevents a missing constant or a mistaken interpretation.

Imagine a water-treatment tank with a net flow rate $r(t)$ in $\frac{\mathrm{L}}{\mathrm{min}}$. The quantity $\int_0^{30}r(t)\,dt$ has liters as its units and describes net change in water volume. It is not automatically the total water present, because that additionally requires an initial volume. The narrative gives a reason to care about bounds, signs, and units before applying a rule.

Use linearity and the power rule

Integration respects sums and constant multiples: $\int[af(x)+bg(x)]\,dx=a\int f(x)\,dx+b\int g(x)\,dx$. This lets a polynomial be integrated term by term. For $n\ne-1$, the reverse-power rule is $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$. The $+C$ represents all vertical shifts with the same derivative.

The excluded case $n=-1$ is important: $\int x^{-1}\,dx=\ln|x|+C$, not division by zero in the power formula. The absolute value preserves both positive and negative parts of the nonzero domain. This exception is a structural signal that a logarithm, not a power, is the appropriate antiderivative.

Recognize reverse Chain Rule structure

Substitution is useful when the integrand contains a function and, up to a constant, its derivative. If $u=g(x)$ and $du=g'(x)\,dx$, then $\int f(g(x))g'(x)\,dx=\int f(u)\,du$. It is a change of variable that follows the same dependency structure as the Chain Rule.

For the tank, an inflow model such as $r(t)=2t(t^2+1)^4$ contains the inner quantity $t^2+1$ and its derivative $2t$. Set $u=t^2+1$, integrate $u^4$, then return to $t$. Do not use substitution just because parentheses appear; identify the matched derivative explicitly.

Evaluate definite integrals with meaning

If $F'(t)=r(t)$, then $\int_a^b r(t)\,dt=F(b)-F(a)$. Apply the upper input and subtract the lower input; the order reflects accumulated change from $a$ to $b$. Negative portions of a rate graph subtract from net change, so a result of zero can describe genuine movement in both directions.

Return to the tank: $V(30)=V(0)+\int_0^{30}r(t)\,dt$. If the problem asks for total inflow rather than net volume change, negative outflow portions must be handled separately or with an absolute-value model. Interpretation comes after evaluation, but it should guide setup from the beginning.

Narrative challenge: update the tank

A tank starts with $40\,\mathrm{L}$ and has net rate $r(t)=6-0.2t\,\frac{\mathrm{L}}{\mathrm{min}}$ for $0\le t\le10\,\mathrm{min}$. Write, but do not initially evaluate, an expression for the final volume. Then decide whether the integral is net change or total inflow.

Show a solution path

The final volume is $V(10)=40\,\mathrm{L}+\int_0^{10}(6-0.2t)\,dt$. The integral measures net volume change because $r(t)$ is signed. On this interval the rate remains positive, so it also happens to equal total inflow, but that conclusion comes from checking the rate rather than from the word “integral.”