Finding Areas
\(\DeclareMathOperator{cosec}{cosec}\)
Equation
\[R=\int ^b_{a}f(x) \, dx-\int ^b_{a}g(x) \, dx =\int ^b_{a}(f(x)-g(x)) \, dx \]
\(f(x)=\sin2x\), \(g(x)=\sin x\cos^2x\)
- Integrate between \(0\) and \(\frac{\pi}{2}\)
-
\[\int _{0}^{\frac{\pi}{2}}(\sin2x-\sin x\cos^2x) \, dx=\left[ \frac{1}{2}\cos2x+\frac{1}{3}\cos^3x \right]^{\frac{\pi}{2}}_{0}\]
Differentiating Parametric Equations
\[A=\int y \, dx=\int y\frac{dx}{dt} \, dt \]
Parametric example, \(x=t(1+t)\), \(y=\frac{1}{1+t}\)
- Find the indefinite integral:
\(\(x=t(1+t)=t+t^2\)\)
- \(\(\frac{dx}{dt}=1+2t\)\)
- $$\text{Area}=\int \frac{1}{1+t}\times(1+2t) \, dt $$
- Find limits:
\(\(x=2t(t+1)=2\)\)
- \(\(t^2+t-2=0\)\)
- \(\(\cancel{ t=-2 } \text{ or } t=1\)\)
- Substitute limits in: \(\(\int _{0}^1 \frac{1+2t}{1+t} \, dt\)\)
- $$\int _{0}^1\left( 2-\frac{1}{1+t} \right) \, dt $$
- \(\(\left[2t-\ln |1+t|te\right]^1_{0}\)\)
The diagram with curve \(C\) with parametric equations: \(x=3t^2\), \(y=\sin2t\), \(t\geq 0\)
- Find the roots of this curve
- \(\sin2t=0\)
- \(t=0\), \(t=\frac{\pi}{2}\)
- \(\frac{dx}{dt}=6t\)
- \(\therefore\int \sin2t \, dx=\int \sin2t\times6t \, dt\)
- Use integration by parts
- \(u=6t\)
- \(\frac{du}{dx}=6\)
- \(\frac{dv}{dx}=\sin2t\)
- \(v=-\frac{1}{2}\cos2t\)
- \(-3t\cos2t-\int (-3\cos2t) \, dx\)
- \(=-3t\cos2t-\frac{3}{2}\sin2t\)
- \(-\frac{3\pi}{2}\cos \pi\cancel{ -\frac{3}{2}\sin \pi }\)
- \(=\frac{3}{2}\pi\)