Let
𝑓
be a differentiable function with domain
𝑅
.

We define
𝐹
(
𝑥
)
=
∫
0
𝑥
𝑓
(
𝑡
)
𝑑
𝑡
.

Given the following table:

x f(x) F(x)
1 4 3
2 1 4
3 2 3
4 3 1
5 4 2

Compute
∫
3
5
(
𝑥
−
2
)
𝑓
′
(
𝑥
)
𝑑
𝑥
. Hint: use integration by parts and FTC part 2.

Enter your answer in decimal form. Round to two decimal places if needed (e.g. enter 0.1 as 0.1, enter 0.2345 as 0.23 or 0.24).

Question

Let 
𝑓
 be a differentiable function with domain 
𝑅
 .

We define 
𝐹
(
𝑥
)
=
∫
0
𝑥
𝑓
(
𝑡
)
𝑑
𝑡
.

Given the following table:

x	f(x)	F(x)
1	4	3
2	1	4
3	2	3
4	3	1
5	4	2

Compute 
∫
3
5
(
𝑥
−
2
)
𝑓
′
(
𝑥
)
𝑑
𝑥
 . Hint: use integration by parts and FTC part 2.

Enter your answer in decimal form. Round to two decimal places if needed (e.g. enter 0.1 as 0.1, enter 0.2345 as 0.23 or 0.24).

BlackTom AI · Accepted Answer

We need to evaluate the integral ∫ from 3 to 5 of (x−2) f′(x) dx. Start by applying integration by parts. Let u = x−2, dv = f′(x) dx. Then du = dx and v = f(x). The integration by parts formula gives:

∫ u dv = uv|_a^b − ∫ v du.

Here, ∫_3^5 (x−2) f′(x) dx = (x−2) f(x) |_3^5 − ∫_3^5 f(x) dx.

Compute the boundary term (x−2) f(x) at the endpoints using the provided table: at x = 5, (5−2) f(5) = 3 × 4 = 12; at x = 3, (3−2) f(3) = 1 × 2 = 2. Therefore, (x−2) f(x)|_3^5 = 12 − 2 = 10.

Next, evaluate ∫_3^5 f(x) dx using F as the antiderivative of f, i.e., F′ = f and F(x) = ∫_0^x f(t) dt. Thus ∫_3^5 f(x) dx = F(5) − F(3). The table gives F(5) = 2 and F(3) = 3, so F(5) − F(3) = 2 − 3 = −1.

Putting it together: ∫_3^5 (x−2) f′(x) dx = [ (x−2) f(x) ]_3^5 − ∫_3^5 f(x) dx = 10 − (−1) = 11.

Since the instruction asks for decimal form, we express this as 11.00. Therefore, the computed value is 11.00.

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