AP Calculus BC - Integral

Question

If ∫₀⁶ f(x) dx = 15, ∫₀⁴ (f(x)/3) dx = -3, find ∫₆⁴ (f(x)+3)dx

Answer 1

Answer 2

If ₀∫⁴ (f(x)/3) dx = -3 the rules of integrals allow us to bring out the constant to become:

(1/3)₀∫⁴ f(x) dx = -3

Then multiply both sides by 3 to get:

₀∫⁴ f(x) dx = -9

We know that:

₀∫⁶ f(x) dx = ₀∫⁴ f(x) dx + ₄∫⁶ f(x) dx therefore:

15 = -9 + ₄∫⁶ f(x) dx or

₄∫⁶ f(x) dx = 24 (adding 9 to both sides)

We know that:

_a∫^b f(x) dx = - _b∫^a f(x) dx

Therefore ₆∫⁴ f(x) dx = -24

We also know that ₆∫⁴ f(x) + 3 dx = ₆∫⁴ f(x) dx + ₆∫⁴ 3 dx

We can evaluate ₆∫⁴ 3 dx using the 2nd Fundamental Theorem of Calculus

₆∫⁴ 3 dx = 3x evaluated between 6 and 4:

3(4) - 3(6) = 12 - 18 = -6

So ₆∫⁴ f(x) + 3 dx = -24 + -6 = -30