Câu hỏi
Prob 12. Suppose f(x) is differentiable, f(1)=3,f(3)=1 and int _(1)^3xf'(x)dx=13 What is the average value of f on the interval [1,3]
Giải pháp
4.5
(267 Phiếu)
Đức An
chuyên gia · Hướng dẫn 6 năm
Trả lời
Let the average value of
on the interval
be denoted by
. Then
A = \frac{1}{3-1} \int_1^3 f(x) dx = \frac{1}{2} \int_1^3 f(x) dx
We use integration by parts to evaluate
. Let
and
. Then
and
.Using integration by parts, we have:
\int_1^3 xf'(x) dx = xf(x) \Big|_1^3 - \int_1^3 f(x) dx = 3f(3) - 1f(1) - \int_1^3 f(x) dx
We are given that
,
, and
. Substituting these values, we get:
13 = 3(1) - 1(3) - \int_1^3 f(x) dx
13 = 0 - \int_1^3 f(x) dx
\int_1^3 f(x) dx = -13
Therefore, the average value of
on
is:
A = \frac{1}{2} \int_1^3 f(x) dx = \frac{1}{2}(-13) = -\frac{13}{2}
Thus, the average value of
on the interval
is
.