Question

If
`\int\limits^x_a` f(t)dt=3x^2+12x, what are the possible values of a? Verify that your values of a are correct.

Answer 1

a = 0, -4

Explanation:

Integral is 3t² + 12t

[3x² + 12x] - [3a² + 12a]

= 3x² + 12x

There's no constant in the integral:

3a² + 12a = 0

a² + 4a = 0

a(a + 4) = 0

a = 0, -4

Verification

Integral is 3t² + 12t

a = -4

[3x² + 12x] - [3(-4)² + 12(-4)]

3x² + 12x

a = 0

[3x² + 12x] - [3(0)² + 12(0)]

3x² + 12x