Let f be twice differentiable with f(0)=12, f(3)=9, and f '(3)=4.Evaluate the integral upper limit 3 and lower integral 0 xf'' (x)dx
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using integration by parts,
∫xf" dx = xf' - ∫f' dx = (xf' - f)[0,3]
= [3f'(3)-f(3)]-[0f'(0)-f(0)]
= (3*4-9)-(0-12)
= 3+12 = 15
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To evaluate the integral ∫₀³ x⋅f''(x)dx, we can use integration by parts twice. Here's how you can solve it step by step:
Step 1: Start by using integration by parts for the first time. The formula for integration by parts is ∫u⋅dv = u⋅v - ∫v⋅du.
In this case, let u = x and dv = f''(x)dx. This means that du = dx and v = ∫f''(x)dx.
Step 2: Integrate dv to find v. Since f''(x) represents the second derivative of f(x), integrating it once gives us the first derivative of f(x), denoted as f'(x). So, v = f'(x).
Step 3: Now, we can apply the integration by parts formula: ∫u⋅dv = u⋅v - ∫v⋅du. Plugging in the values, we have:
∫₀³ x⋅f''(x)dx = x⋅f'(x) - ∫₀³ f'(x)dx.
Step 4: We will use integration by parts again for the second integral. Let u = f'(x) and dv = dx. Thus, du = f''(x)dx and v = ∫dx = x.
Step 5: Applying the integration by parts formula to the second integral, we get:
∫₀³ f'(x)dx = f'(x)⋅x - ∫₀³ x⋅f''(x)dx.
Step 6: Rearrange the equation from Step 5 to solve for the original integral:
∫₀³ x⋅f''(x)dx = x⋅f'(x) - ∫₀³ f'(x)dx.
Step 7: We know that f(3) = 9, so plugging this into the equation gives us:
∫₀³ x⋅f''(x)dx = 3⋅f'(3) - ∫₀³ f'(x)dx.
Step 8: We are given that f'(3) = 4, so we can substitute this into the equation:
∫₀³ x⋅f''(x)dx = 3⋅4 - ∫₀³ f'(x)dx.
Step 9: We need to find the integral of f'(x) from x = 0 to x = 3. However, we don't have the function f(x) or f'(x) explicitly. Instead, we have some initial conditions.
Step 10: Since f(0) = 12, we can calculate the value of the definite integral:
∫₀³ f'(x)dx = f(3) - f(0) = 9 - 12 = -3.
Step 11: Plugging this into the equation from Step 8, we have:
∫₀³ x⋅f''(x)dx = 3⋅4 - (-3) = 12 + 3 = 15.
Therefore, the value of the integral ∫₀³ x⋅f''(x)dx is 15.