Suppose

5
∫ f(x)dx=6

2

3
∫ f(x)dx=1

2

5
∫ f(x)dx=4

4

4
∫ (6f(x)−1)dx= ?

3

help, please? I got the answer of 5, but it says its wrong

To find the integral of a function, you need to use the fundamental theorem of calculus, which states that if F(x) is an antiderivative of f(x), then ∫ f(x) dx from a to b is equal to F(b) - F(a). In other words, we can evaluate the integral by subtracting the value of the antiderivative at the upper limit (b) from the value at the lower limit (a).

Let's look at the given information step by step:

1. ∫f(x) dx from 5 to 2 = 6
2. ∫f(x) dx from 3 to 2 = 1
3. ∫f(x) dx from 5 to 4 = 4

You are asked to evaluate the integral ∫(6f(x) - 1) dx from 4 to 3.

To solve this, let's break it down:

∫(6f(x) - 1) dx from 4 to 3
= (6 * ∫f(x) dx - ∫1 dx) from 4 to 3

Now, using the fundamental theorem of calculus, we can evaluate each integral:

= (6 * (F(x))] from 4 to 3 - (x) from 4 to 3
= 6 * [F(3) - F(4)] - [3 - 4]
= 6 * [∫f(x) dx from 3 to 2 - ∫f(x) dx from 5 to 4] + 1
= 6 * (1 - 4) + 1
= -18 + 1
= -17

Therefore, the correct answer to ∫(6f(x) - 1) dx from 4 to 3 is -17, not 5.