Answer:-12
Step-by-step explanation:To evaluate the integral ∫₃₅ x f ′′(x) dx, we can use integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
Let's denote u = x and dv = f ′′(x) dx. Then, we can differentiate u to get du = dx and integrate dv to get v = ∫ f ′′(x) dx.
Given that f ′′(x) is continuous, we can integrate it twice to get f(x), and we are given f(3), f'(3), f''(3), f(5), f'(5), and f''(5). We can use these values to find the constants of integration.
First, integrate f ′′(x) once to get f ′(x):
∫ f ′′(x) dx = f ′(x) + C₁
Using the given information, we have:
f ′(3) = 4
∫ f ′′(x) dx from 3 to 3 = f ′(3) - f ′(3) = 0
So, C₁ = 4.
Next, integrate f ′(x) to get f(x):
∫ f ′(x) dx = f(x) + C₂
Using the given information:
f(3) = 3
∫ f ′(x) dx from 3 to 3 = f(3) - f(3) = 0
So, C₂ = 3.
Now, we can integrate by parts using the values of u, v, du, and dv:
∫₃₅ x f ′′(x) dx = uv - ∫ v du
= x ∫ f ′′(x) dx - ∫ (∫ f ′′(x) dx) dx
Substitute the values of u and v, and integrate v with respect to x:
∫₃₅ x f ′′(x) dx = x(f ′(x) + 4) - ∫ (f ′(x) + 4) dx
= x(f ′(x) + 4) - ∫ f ′(x) dx - ∫ 4 dx
Integrate f ′(x) using the previous result and integrate the constant term:
∫₃₅ x f ′′(x) dx = x(f(x) + 3) - (f(x) + 3x) - 4x
Now, plug in the values for x = 5 and x = 3, and use the given information to evaluate the expression:
∫₃₅ x f ′′(x) dx = 5(f(5) + 3) - (f(5) + 3(5)) - 4(5) - [3(f(3) + 3) - (f(3) + 3(3)) - 4(3)]
= 5(2 + 3) - (2 + 15) - 20 - [3(3 + 3) - (3 + 9) - 12]
= 25 - 17 - 20 - [18 - 12 - 12]
= -12
Therefore, ∫₃₅ x f ′′(x) dx = -12.