If
="absmiddle" class="latex-formula"> f(t)dt=3x^2+12x, what are the possible values of a? Verify that your values of a are correct.
1 answer:
Answer:
a = 0, -4
Step-by-step 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
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