Integral Topic Assessment Answers

| Mistake | Check | |---------|-------| | Forgetting +C | Always add after integrating indefinite | | Wrong u‑sub (missing du) | Differentiate your u – do you see that factor? | | Sign errors in by‑parts | Check formula: ∫u dv = uv – ∫v du | | Arithmetic with bounds | Substitute carefully, simplify step by step | | Not checking improper integrals | Evaluate limit, check for divergence |

Therefore, the area between the curves is 4/3.

$$∫[0,2] (2x + 1)dx = [x^2 + x]_0^2 = (2^2 + 2) - (0^2 + 0) = 6$$

Answer: [x^2 + x] from 0 to 2 = (2^2 + 2) - (0^2 + 0) = 6

Remember to subtract the "bottom" function from the "top" function.

y = x² and y = x from 0 to 1 ∫₀¹ (x – x²) dx = 1/6

The area between the curves is given by:

= x^2 e^x - 2∫x e^x dx

Write down your substitution or parts choice clearly. Example (by parts): ∫ x·eˣ dx → u = x, dv = eˣ dx → du = dx, v = eˣ → = x·eˣ – ∫ eˣ dx = x·eˣ – eˣ + C.