# Tag Archives: Integration Example 1 Find the derivative of $\sin (2x-5)$ and use this result to deduce $\displaystyle \int 10 \cos (2x-5) dx$. \begin{align} \displaystyle \frac{d}{dx} \sin (2x-5) &= \cos (2x-5) \times \frac{d}{dx} (2x-5) &\color{green}{\text{Chain Rule}} \\ &= \cos (2x-5) \times 2 \\ &= 2 \cos (2x-5) \\ \sin (2x-5) &= \int 2 […] # Volumes for Two Functions If the region bounded by the upper function y_{upper}=f(x) and the lower funciton y_{lower}=g(x), and the lines x=a and x=b is revolved about the x-axis, then its volume of revolution is given by:  \begin{align} \displaystyle V &= \int_{a}^{b}{\Big([f(x)]^2 – [g(x)]^2\Big)}dx \\ &= \int_{a}^{b}{\Big(y_{upper}^2 – y_{lower}^2\Big)}dx \end{align}  Example 1 Find the volume of revolution […] # Volumes using Integration Volume of Revolution We can use integration to find volumes of revolution between x=a and x=b. When the region enclosed by y=f(x), the x-axis, and the vertical lines x=a and x=b is revolved through 2 \pi or 360^{\circ}about the x-axis to generate a solid, the volume of the solid is given by:  \begin{align} \displaystyle […] # Kinematics using Integration Distances from Velocity Graphs Suppose a car travels at a constant positive velocity 80 \text{ km h}^{-1} for 2 hours.  \begin{align} \displaystyle \text{distance travelled} &= \text{speed} \times \text{time} \\ &= 80 \text{ km h}^{-1} \times 2 \text{ h} \\ &= 160 \text{ km} \end{align}  We we sketch the graph velocity against time, the […] # Area Between Two Functions If two functions f(x) and g(x) intersect at x=1 and x=3, and f(x) \ge g(x) for all 1 \le x \le 3, then the area of the shaded region between their points of intersection is given by:  \begin{align} \displaystyle A &= \int_{1}^{3}{f(x)}dx – \int_{1}^{3}{g(x)}dx \\ &= \int_{1}^{3}{\Big[f(x)-g(x)\Big]}dx \end{align}  Example 1 Find the area […] # Definite Integration by Substitution  \begin{align} \displaystyle u &= f(x) \\ du &= \dfrac{du}{dx} \times dx \\ \end{align} Converting x-values to corresponding u-values are required. Example 1 Find \displaystyle 2\int_{0}^{1}{\sqrt{2x+1}}dx. \( \begin{align} \displaystyle \text{Let } u &= 2x+1 \\ \dfrac{du}{dx} &= 2 \\ du &= 2dx \\ u &= 2 \times 1 + 1 = 3 &x=1\\ u &= […] # Definite Integration of Power Functions \displaystyle \int_{n}^{m}{(ax+b)^k}dx = \dfrac{1}{a(k+1)}\Big[(ax+b)^{k+1}\Big]_{n}^{m}+c Example 1 Find \displaystyle \int_{0}^{1}{(2x+1)^5}dx. \( \begin{align} \displaystyle \int{(2x+1)^5}dx &= \dfrac{(2x+1)^{5+1}}{2(5+1)} \\ &= \dfrac{1}{12}\big[(2x+1)^{6}\big]_{0}^{1} \\ &= \dfrac{1}{12}\big[(2 \times 1+1)^{6} – (2 \times 0+1)^{6}\big] \\ &= \dfrac{1}{12}(729 – 1) \\ &= \dfrac{728}{12} \\ &= \dfrac{182}{3} \end{align} Example 2 Find $\displaystyle \int_{0}^{1}{\dfrac{1}{(3x-2)^4}}dx$. \begin{align} \displaystyle \int_{0}^{1}{\dfrac{1}{(3x-2)^4}}dx &= \int_{0}^{1}{(3x-2)^{-4}}dx \\ &= \bigg[\dfrac{(3x-2)^{-4+1}}{3(-4+1)}\bigg]_{0}^{1} \\ […] # Definite Integral of Rational Functions  \begin{align} \displaystyle \int_{n}^{m}{\dfrac{1}{x}}dx &= \big[\log_e{x}\big]_{n}^{m} \\ &= \log_{e}{m} – \log_{e}{n} \end{align}  Generally,  \begin{align} \displaystyle \int_{n}^{m}{\dfrac{f'(x)}{f(x)}}dx &= \big[\log_e{f(x)}\big]_{n}^{m} \\ &= \log_{e}{f(m)} – \log_{e}{f(n)} \end{align}  Example 1 Find \displaystyle \int_{1}^{5}{\dfrac{2}{x}}dx. \( \begin{align} \displaystyle \int_{1}^{5}{\dfrac{2}{x}}dx &= 2\int_{1}^{5}{\dfrac{1}{x}}dx \\ &= 2\big[\log_{e}{x}\big]_{1}^{5} \\ &= 2 \log_{e}{5} – 2 \log_{e}{1} \\ &= 2 \log_{e}{5} – 2 \times […] # Definite Integral of Exponential Functions  \begin{align} \displaystyle \int_{n}^{m}{e^{ax+b}}dx &= \dfrac{1}{a}\big[e^{ax+b}\big]_{n}^{m} \\ &= \dfrac{1}{a}\big[e^{am+b}-e^{an+b}\big] \\ \end{align}  Example 1 Find \displaystyle \int_{2}^{4}{e^{2x-4}}dx, leaving the answer in exact form. \( \begin{align} \displaystyle \int_{2}^{4}{e^{2x-4}}dx &= \dfrac{1}{2}\big[e^{2x-4}\big]_{2}^{4} \\ &= \dfrac{1}{2}\big[e^{2 \times 4 – 4} – e^{2 \times 2 – 4}\big] \\ &= \dfrac{e^4 – e^{0}}{2} \\ &= \dfrac{e^4 – 1}{2} \\ \end{align} […] # Definite Integrals

The Fundamental Theorem of Calculus For a continuous function $f(x)$ with antiderivative $F(x)$, $$\displaystyle \int_{a}^{b}{f(x)}dx = F(b) – F(a)$$ Properties of Definite Integrals The following properties of definite integrals can all be deductefd from the fundamental theorem of calculus: $\displaystyle \int_{a}^{a}{f(x)}dx = 0$ $\displaystyle \int_{b}^{a}{f(x)}dx = -\int_{a}^{b}{f(x)}dx$ $\displaystyle \int_{b}^{a}{f(x)}dx + \int_{c}^{b}{f(x)}dx = \int_{c}^{a}{f(x)}dx$ \$\displaystyle \int_{b}^{a}{\big[f(x) […]