Exercise Set 12
Determine the following indefinite integrals
- \(\int 2x^7\; dx\)
- \(\int 5x^2+2x-1 \; dx\)
- \(\int \sin(\theta)+\cos(\theta) \; d\theta\)
- \(\int 3e^t + \frac{2}{t} \; d t\)
- \(\int x^\frac{1}{2} \; dx\)
Determine the following definite integrals
- \(\int_0^2 x^2-2x\; dx\)
- \(\int_4^{16} \frac{2}{\sqrt{x}} \; dx\)
- \(\int_{-r}^r \frac{mx^2}{2r}\: dx\)
- \(\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\cos(x)\; dx\)
- \(\int_{\ln(2)}^{\ln(3)} e^x\; dx\)
Let \(F(x)\) be an anti-derivative of \(f(x)\). Find an anti-derivative of \(f(ax+b)\) where \(a\) and \(b\) are constants. Hint: use the chain rule on \(F(ax+b)\).
Use your result to the previous question to find the following indefinite integrals
- \(\int (2x+6)^2\; dx\)
- \(\int (2x+6)^{99} \; dx\)
- \(\int \sin(3x+2) \; dx\)
- \(\int \cos(-4x-1) \; dx\)
- \(\int \sec^2(9x+2) \; dx\)
- \(\int \frac{1}{2x+7}\; dx\)
- \(\int e^{-3x+1}\; dx\)
- \(\int \frac{1}{(5x+1)^2}\; dx\)
- \(\int \sqrt{3x+4}\; dx\)
Find the area between the curve and the \(x\)-axis of the function \(f(x)=x^3-x\) and the lines \(x=0\) and \(x=2\).
The velocity \(v\) (\(\text{ms}^{-1}\)) of a particle moving in a straight line at time \(t\) (\(\text{s}\)) is given by: \[v=6t^2-30t-36\quad t>0\] By integrating, find the distance travelled by the particle between \(t=3\) and \(t=5\) seconds.
The Root Mean Square (RMS) value of a signal over the interval \(a\) to \(b\) is defined as \[\left(\frac{1}{b-a}\int_{a}^{b} (f(x))^2 \; dx\right)^\frac{1}{2}\] Find the RMS of the voltage signal \[V(t)=3\sin(2t)+5\cos(4t)\] between \(a=0\) and \(b=2\pi\).
The volume of revolution of a curve is defined as \[\int_a^b \pi (f(x))^2 \; dx\] Find the volume of revolution of the curve \[f(x)=e^{-2x}\] between \(a=1\) and \(b=2\).
The points \((1,1)\) and \((2,8)\) on the curve \(y=x^3\) are joined by a straight line. Calculate the area between the line and the curve.