Suppose the integral from 2 to 6 of g of x, dx equals 12 and the integral from 5 to 6 of g of x, dx equals negative 3 , find the value of the integral from 2 to 5 of 3 times g of x, dx

INT g(x)dx from 2 to 6=

INT g(x)dx from 2 to 5 + INT(g(x) from 5 to 6

12=Int(g(x) from 2 to 5 -3
so INT gIx)dx from 2 to 5 = 15
so three times int(g)x from 2 to 5 is then 3*15

To find the value of the integral from 2 to 5 of 3 times g(x) dx, we can use the properties of integrals and apply some algebraic manipulation.

First, let's use the linearity property of integrals, which states that the integral of the sum or difference of two functions is equal to the sum or difference of the integrals of each function separately. Using this property, we can rewrite the integral we need to find as follows:

∫[2 to 5] (3 * g(x)) dx = 3 * ∫[2 to 5] g(x) dx

Next, let's use the property of integrals that states if we take the integral of a function in a smaller interval within the range of integration of a larger integral, the value of the integral will be the same. We can use this property to rewrite the integral as follows:

3 * ∫[2 to 5] g(x) dx = 3 * (∫[2 to 6] g(x) dx - ∫[5 to 6] g(x) dx)

Substituting the given values, we have:

3 * (∫[2 to 6] g(x) dx - ∫[5 to 6] g(x) dx) = 3 * (12 - (-3))

Simplifying further:

3 * (12 - (-3)) = 3 * (12 + 3) = 3 * 15 = 45

Therefore, the value of the integral from 2 to 5 of 3 times g(x) dx is 45.