Answer: \(\int\frac{1}{\sqrt{3x+2}}\,\mathrm dx=\frac23(3x+2)^{1/2}+c\), and \(\int_{0.5}^{a}\mathrm e^{1-2x}\,\mathrm dx=\frac12-\frac12\mathrm e^{1-2a}\).
Set up the integral using the correct limits or antiderivative first. When definite integration is used, evaluate the upper limit and lower limit separately before subtracting.
For part (a),
\(\int(3x+2)^{-1/2}\,\mathrm dx=\frac{(3x+2)^{1/2}}{\frac12\cdot3}+c=\frac23(3x+2)^{1/2}+c.\)
For part (b),
\(\int\mathrm e^{1-2x}\,\mathrm dx=-\frac12\mathrm e^{1-2x}+c.\)
Therefore
\(\int_{0.5}^{a}\mathrm e^{1-2x}\,\mathrm dx=\left[-\frac12\mathrm e^{1-2x}\right]_{0.5}^{a}\).
This gives
\(-\frac12\mathrm e^{1-2a}+\frac12\mathrm e^{0}=\frac12-\frac12\mathrm e^{1-2a}.\)
This gives the required answer: \(\int\frac{1}{\sqrt{3x+2}}\,\mathrm dx=\frac23(3x+2)^{1/2}+c\), and \(\int_{0.5}^{a}\mathrm e^{1-2x}\,\mathrm dx=\frac12-\frac12\mathrm e^{1-2a}\).
