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
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To find the value of the integral from 2 to 5 of 3 times g of x, dx, we can use the linearity property of integrals:
∫[a to b] (c*f(x)) dx = c * ∫[a to b] f(x) dx
In this case, we have:
∫[2 to 5] 3 * g(x) dx
Since we're given the integrals of g(x) from 2 to 6 and from 5 to 6, we can split the integral in question into two parts:
∫[2 to 5] 3 * g(x) dx = ∫[2 to 5] 3 * g(x) dx + ∫[5 to 6] 3 * g(x) dx
We'll consider each part separately.
For the first part (∫[2 to 5] 3 * g(x) dx), we can apply the linearity property to obtain:
∫[2 to 5] 3 * g(x) dx = 3 * ∫[2 to 5] g(x) dx
Since we know that ∫[2 to 6] g(x) dx = 12, we can express this integral in terms of known values:
∫[2 to 5] g(x) dx = ∫[2 to 6] g(x) dx - ∫[5 to 6] g(x) dx
Plugging in the given values, we get:
∫[2 to 5] g(x) dx = 12 - (-3)
∫[2 to 5] g(x) dx = 12 + 3
∫[2 to 5] g(x) dx = 15
Now, substituting this value back into the previous expression:
∫[2 to 5] 3 * g(x) dx = 3 * 15
∫[2 to 5] 3 * g(x) dx = 45
Therefore, the value of the integral from 2 to 5 of 3 times g of x, dx is 45.