If g(1)=4, g(5)=5, and ∫ [1,5] g(x) dx =-8, how do you evaluate the integral ∫ [1,5] x g’(x) dx?
∫ x g'(x) dx = ∫ x dg
so, using integration by parts, that is
∫ x dg = (xg - ∫ g dx ) [1..5] = (5g(5) - 1g(1)) - (-8) = 25-4+8 = 29
Well, let's give it a shot! But first, let's take a moment to appreciate the fact that calculus can really integrate people's lives and problems. Now, to evaluate the integral ∫ [1,5] x g’(x) dx, let's use integration by parts.
Using the product rule for differentiation, we know that (x * g(x))' = x * g'(x) + g(x).
Now, let's integrate both sides of the equation:
∫ [(x * g(x))'] dx = ∫ [(x * g'(x) + g(x))] dx.
Applying the technique of integration by parts, we have:
(x * g(x)) = ∫ [x * g'(x)] dx + ∫ [g(x)] dx.
This can be rewritten as:
∫ [x * g'(x)] dx = (x * g(x)) - ∫ [g(x)] dx.
Now, let's plug in the bounds of integration:
∫ [1,5] x * g'(x) dx = (5 * g(5)) - ∫ [1,5] g(x) dx.
Since we know g(5) = 5 and ∫ [1,5] g(x) dx = -8, we can substitute these values:
∫ [1,5] x * g'(x) dx = (5 * 5) - (-8) = 25 + 8 = 33.
So, the value of the integral ∫ [1,5] x * g'(x) dx is 33. Now we can celebrate the fact that in this case, math and humor have integrated perfectly!
To evaluate the integral ∫ [1,5] x g’(x) dx, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's assign u = x and dv = g'(x)dx.
To find du, differentiate u with respect to x:
du = dx
To find v, integrate dv:
∫ dv = ∫ g'(x) dx = g(x)
Now we have all the components to apply the formula for integration by parts:
∫ [1,5] x g’(x) dx = xg(x) - ∫ [1,5] g(x) dx
Since we have g(1) = 4 and g(5) = 5, we can evaluate the expression:
∫ [1,5] x g’(x) dx = 5g(5) - 1g(1) - ∫ [1,5] g(x) dx
Plugging in the given values:
∫ [1,5] x g’(x) dx = 5(5) - 1(4) - (-8)
∫ [1,5] x g’(x) dx = 25 - 4 + 8
∫ [1,5] x g’(x) dx = 29
Therefore, the integral ∫ [1,5] x g’(x) dx evaluates to 29.
To evaluate the integral ∫ [1,5] x g’(x) dx, we can use integration by parts. Integration by parts involves using a specific formula:
∫ u dv = uv - ∫ v du
Let's apply this formula to our integral. Let's choose u = x and dv = g'(x) dx. Therefore, du = dx and v = g(x).
Using the formula, we get:
∫ [1,5] x g'(x) dx = xg(x) - ∫ [1,5] g(x) dx
Now, we know that g(1) = 4, g(5) = 5, and ∫ [1,5] g(x) dx = -8. Plugging in these values, we have:
∫ [1,5] x g'(x) dx = xg(x) - (-8)
= xg(x) + 8
To find the value of the integral, we need to determine the value of xg(x) for x = 5 and x = 1.
For x = 5: xg(x) = 5 * g(5) = 5 * 5 = 25
For x = 1: xg(x) = 1 * g(1) = 1 * 4 = 4
Therefore, the integral ∫ [1,5] x g'(x) dx equals 25 + 8 = 33.