Introduction to integration; boundary condition; areas practice questions.

Determine ∫ (6x² − 4x + 1) dx.

• 6x³ − 4x² + x + C
• 2x³ − 2x² + x + C
• 12x − 4 + C
• 2x³ − 4x² + x + C

Answer: 2x³ − 2x² + x + C

Raise each power by one and divide by the new power: 6x³/3 − 4x²/2 + x = 2x³ − 2x² + x, plus the constant of integration.

Calculate the exact value of ∫₀² 3x² dx.

• 12
• 24
• 6
• 8

Answer: 8

An antiderivative is x³, so the integral equals 2³ − 0³ = 8. Substituting into 3x² instead of the antiderivative is a common error.

The curve y = 4 − x² crosses the x-axis at x = −2 and x = 2. Determine the exact area enclosed between the curve and the x-axis.

• 16/3
• 8
• 32/3
• 64/3

Answer: 32/3

Area = ∫ from −2 to 2 of (4 − x²) dx = [4x − x³/3] = (8 − 8/3) − (−8 + 8/3) = 32/3. Forgetting the negative limit's contribution gives 16/3.

Determine ∫ 4x⁻³ dx.

• 2/x² + C
• −12/x⁴ + C
• −2/x² + C
• −4/(3x²) + C

Answer: −2/x² + C

∫ 4x⁻³ dx = 4·x⁻²/(−2) + C = −2x⁻² + C = −2/x² + C. Dropping the negative from the new power −2 is the classic sign error.

The curves y = x and y = x² intersect at x = 0 and x = 1. Determine the exact area of the region enclosed between them.

• 1/3
• 1/6
• 1/2
• 5/6

Answer: 1/6

On [0, 1] the line lies above the parabola, so Area = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.

Given that ∫₁ᵏ 2x dx = 15 with k > 1, determine the value of k.

• 16
• √15
• 4
• 8

Answer: 4

The integral equals k² − 1, so k² − 1 = 15 gives k² = 16 and hence k = 4 (taking k > 1). Forgetting the lower limit leads to k = √15.