HSC Extension 1 and 2 Mathematics/Integration

Area

• Fundamental Theorem of Calculus: $\int _{a}^{b}f(x)dx=F(b)-F(a)$, where ${\frac {d}{dx}}F(x)=f(x)$

Volume of solids of revolution

Recall that the volume of a solid can be found by $V=Ad\$ where $A$ is the cross-sectional area and $d\$ is the depth of the solid, which is perpendicular to the cross-sectional area.

Similarly, the volume of solids with circular cross sections can be calculated by

• rotating a curve about an axis (generally $x\$ or $y\$ axis)
• integrating to sum the areas of the slices of circles

Since the area of a circle is $A=\pi r^{2}\$, then the integral to evaluate the volume of a solid generated by revolving it around the $x$-axis is $V=\pi \int _{a}^{b}y^{2}dx$

Notice this is a sum of areas of the "slices" of circular cross sections of the solid, i.e. $\sum \pi r^{2}=\pi \sum r^{2}$.

Approximate integration

Trapezoidal rule

• One interval (2 function values): $\int _{a}^{b}f(x)dx\approx {\frac {1}{2}}\times \overbrace {\frac {b-a}{n}} ^{=h}[f(a)+f(b)]$
• $n\$-intervals ($n+1\$ function values): $\int _{a}^{b}f(x)dx\approx {\frac {h}{2}}\left[f(a)+2\sum f(x_{i})+f(b)\right]$

Simpson's rule

$\int _{a}^{b}f(x)dx\approx {\frac {b-a}{6}}\left[f(a)+4f\left({\frac {a+b}{2}}\right)+f(b)\right]$