Introduction to integration; boundary condition; area under curve practice questions.

Find the general antiderivative of f(x) = 6x² + 4x − 5.

• 2x³ + 2x² − 5x + C
• 12x + 4 + C
• 6x³ + 4x² − 5x + C
• 2x³ + 4x² − 5x + C

Answer: 2x³ + 2x² − 5x + C

Integrate term by term: ∫6x² dx = 2x³, ∫4x dx = 2x², ∫−5 dx = −5x, plus the constant C.

A curve has gradient function dy/dx = 4x + 3 and passes through the point (1, 6). Determine the equation of the curve.

• y = 2x² + 3x + 1
• y = 2x² + 3x
• y = 4x² + 3x − 1
• y = 2x² + 3x − 1

Answer: y = 2x² + 3x + 1

Integrating gives y = 2x² + 3x + C; substituting (1, 6): 2 + 3 + C = 6 so C = 1.

Calculate the definite integral of (2x + 1) with respect to x from x = 0 to x = 3.

• 9
• 12
• 10
• 15

Answer: 12

The antiderivative is x² + x; evaluating: (9 + 3) − (0 + 0) = 12.

The area enclosed between the curve y = x², the x-axis and the lines x = 1 and x = 4 is found by evaluating a definite integral. Calculate this area.

• 21
• 64
• 63
• 16

Answer: 21

Area = ∫ from 1 to 4 of x² dx = [x³/3] = 64/3 − 1/3 = 63/3 = 21.

Determine ∫(1/x²) dx, writing the result without negative exponents where possible.

• −1/x + C
• 1/x + C
• ln|x²| + C
• −2/x³ + C

Answer: −1/x + C

Write 1/x² as x⁻²; ∫x⁻² dx = x⁻¹/(−1) = −1/x + C.

Use integration to calculate the area between y = 2x + 1 and the x-axis from x = 0 to x = 2.

• 6
• 5
• 8
• 10

Answer: 6

∫ from 0 to 2 of (2x + 1) dx = [x² + x] = (4 + 2) − 0 = 6.