Answer: \(\frac{\mathrm dy}{\mathrm dx}=\mathrm e^{2x}(1+2x)\); normal \(y-\mathrm e^2=-\frac{1}{3\mathrm e^2}(x-1)\); integral \(\frac32\mathrm e^4+\frac12\).
(a) The curve is
\(y=x\mathrm e^{2x}.\)
Use the product rule:
\(\frac{\mathrm dy}{\mathrm dx}=x(2\mathrm e^{2x})+\mathrm e^{2x}.\)
Therefore
\(\frac{\mathrm dy}{\mathrm dx}=\mathrm e^{2x}(1+2x).\)
(b) At \(x=1\),
\(y=1\cdot\mathrm e^2=\mathrm e^2.\)
Also, the tangent gradient is
\(\mathrm e^2(1+2)=3\mathrm e^2.\)
So the normal gradient is
\(-\frac1{3\mathrm e^2}.\)
The equation of the normal is therefore
\(y-\mathrm e^2=-\frac1{3\mathrm e^2}(x-1).\)
(c) From part (a),
\(\frac{\mathrm d}{\mathrm dx}(x\mathrm e^{2x})=\mathrm e^{2x}+2x\mathrm e^{2x}.\)
So
\(2x\mathrm e^{2x}=\frac{\mathrm d}{\mathrm dx}(x\mathrm e^{2x})-\mathrm e^{2x}.\)
Hence
\(\int_0^2 2x\mathrm e^{2x}\,\mathrm dx =\left[x\mathrm e^{2x}\right]_0^2-\int_0^2\mathrm e^{2x}\,\mathrm dx.\)
Now
\(\left[x\mathrm e^{2x}\right]_0^2=2\mathrm e^4,\)
and
\(\int_0^2\mathrm e^{2x}\,\mathrm dx=\left[\frac12\mathrm e^{2x}\right]_0^2 =\frac12\mathrm e^4-\frac12.\)
Therefore
\(\int_0^2 2x\mathrm e^{2x}\,\mathrm dx =2\mathrm e^4-\left(\frac12\mathrm e^4-\frac12\right) =\frac32\mathrm e^4+\frac12.\)
This completes the solution and gives the required result.
