P4.6 - Integration

Rotating the region between a non-negative curve $y=f(x)$ and the $x$-axis through a full turn produces circular cross-sections. A slice of thickness $dx$ has area $\pi y^2$, so its volume is accumulated by integration.

For $y=x+1$ on $0\le x\le2$, $$V=\pi\int_0^2(x+1)^2\,dx=\pi\left[\frac{(x+1)^3}{3}\right]_0^2=\frac{26\pi}{3}.$$ Square the whole expression for $y$ before integrating, and obtain the limits from the region's $x$-coordinates.

If $x=x(t)$ and $y=y(t)$, replace $dx$ by $x'(t)dt$: $$V=\pi\int_{t_1}^{t_2}y(t)^2x'(t)\,dt.$$ For $x=t^2$, $y=t+1$, $0\le t\le2$, $$V=\pi\int_0^2(t+1)^2(2t)\,dt=\frac{68\pi}{3}.$$ Choose the parameter limits in the direction of increasing $x$; if $x'(t)<0$ throughout, reverse the limits so the volume is positive.

Integration by substitution reverses the chain rule; integration by parts reverses the product rule. First identify which structure is present, then transform every factor, differential and—when the integral is definite—limit consistently.

For $$\int x\sqrt{x^2+4}\,dx,$$ let $u=x^2+4$, so $du=2x\,dx$. Then $$\int x\sqrt{x^2+4}\,dx=\frac12\int u^{1/2}\,du=\frac13(x^2+4)^{3/2}+C.$$ For a definite integral, either change both limits to $u$-values or return to $x$ before applying the original limits—never mix the two.

For $\int x^2e^x\,dx$, take $u=x^2$ and $dv=e^x dx$. Two applications give $$\int x^2e^x\,dx=x^2e^x-2\int xe^x\,dx=e^x(x^2-2x+2)+C.$$ Also, writing $\ln x$ as $1\cdot\ln x$ gives the required result $$\int\ln x\,dx=x\ln x-x+C.$$

When a proper rational expression has factorised linear denominators, decompose it into simpler fractions before integrating. Match a constant numerator to each linear factor and an additional term for every repeated power.

For $$\frac{5x+1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2},$$ matching numerators gives $A=-4$ and $B=9$. Hence $$\int\frac{5x+1}{(x+1)(x+2)}\,dx=-4\ln|x+1|+9\ln|x+2|+C.$$ Differentiate the result to check both coefficients.

A repeated linear factor does not produce another logarithm: $$\int\frac{3}{(x-1)^2}\,dx=-\frac{3}{x-1}+C.$$ Likewise, $$\int\frac{2}{(2x-1)^4}\,dx=-\frac{1}{3(2x-1)^3}+C.$$

A first-order equation is separable when all $y$-dependence can be placed with $dy$ and all $x$-dependence with $dx$. Integrating produces a family of curves—the general solution. An initial or boundary condition then selects a particular member.

For $dy/dx=2xy$ with $y\ne0$, separation gives $$\frac1y\,dy=2x\,dx,$$ so $$\ln|y|=x^2+C\quad\Longrightarrow\quad y=Ae^{x^2}.$$ This non-zero family, together with the equilibrium solution $y=0$ lost when dividing by $y$, is the general solution.

If $y(0)=3$, then $3=Ae^0$, so the particular solution is $$y=3e^{x^2}.$$ Check it by differentiating: $dy/dx=6xe^{x^2}=2x(3e^{x^2})=2xy$, and it satisfies the initial condition.

A reliable sequence is: separate with $dy$ and $dx$ visible; integrate both sides and include one arbitrary constant; simplify only when it helps; use the condition; state the requested form; differentiate to verify. Partial fractions or integration by parts may be needed after separation.

For a parametric curve $x=x(t)$, $y=y(t)$, convert the Cartesian area element $y\,dx$ using $dx=x'(t)dt$. The parameter values at the region's endpoints supply the limits; the curve does not need to be rearranged into $y=f(x)$.

For $x=t^2$, $y=t+1$, $0\le t\le2$, $x'(t)=2t$ and the curve lies above the $x$-axis. Thus $$A=\int_0^2(t+1)(2t)\,dt=\left[\frac23t^3+t^2\right]_0^2=\frac{28}{3}.$$