Hint
Include the minus sign in the remaining integral. The first uv term gives \(\displaystyle \frac{A}{k}\) times \(\displaystyle x^{n}\).
Worked solution
(a)
Choose \(\displaystyle u=\)\(4 x\) and \(\displaystyle dv=\)\(e^{2 x}\) dx. Then \(\displaystyle v=\)\(\displaystyle \frac{e^{2 x}}{2}\) and du=\(4\) dx. Apply \(\displaystyle uv-\int v\) du, obtaining \(\left(2 x - 1\right) e^{2 x}+C\).
Answer: \(\left(2 x - 1\right) e^{2 x} + C\)
(b)
Use upper primitive value minus lower primitive value: \(e^{2}\) − (\(-1\)) = \(1 + e^{2}\).
Answer: \(1 + e^{2}\)
(c)
Subtract the primitive value at the initial abscissa, so the stated initial condition is satisfied.
Answer: \(G(x)=\left(2 x - 1\right) e^{2 x} + 1\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Choose and correctly apply a valid integration method, including inner scaling.
- 1 mark: Give a complete equivalent primitive with an arbitrary constant.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not drop the lower boundary term.
(b) · 2 marks
- 1 mark: Apply the limits to a valid primitive in the correct order.
- 1 mark: Give the complete exact definite-integral value.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: The exponential primitive includes its inner scaling.
(c) · 2 marks
- 1 mark: Find a primitive and determine its constant from the given initial value.
- 1 mark: Give the complete function G satisfying both its derivative and initial condition.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not drop the lower boundary term.
Total: 6 marks.