Another question (basic M2 level)
integrate 2e^x +xe^x
Further question:
Consider a function f(x) being differentiable over the domain [0, +∞) and f(0) = 0. The inverse function of f(x) is g(x). If:
\int_{0}^{f(x)} g(t)dt = x^2 e^x
Find f(x).
Solution:
g[f(x)]f'(x) = 2xe^x + x^2e^x
Since g[f(x)] = x, we have:
xf'(x) = 2xe^x + x^2e^x
For x =/= 0, we have:
f'(x) = 2e^x + xe^x
Integrating, we have:
f(x) = (x+1)e^x + C
Since f(x) is continuous at x=0,
putting limit for x -> 0+,
we have f(0) = 0 and thus C = -1
i.e. f(x) = (x+1)e^x -1
Another question:
Given:
f(x) = \int_{0}^{5x} \frac{sint}{t}dt
and
g(x) = \int_{0}^{sinx} (1+t)^{\frac{1}{t}}dt
Then when x approaches 0, do f(x) and g(x) have the same degree? Explain briefly.
Solution:
Consider
\lim_{x \to 0} \frac{f(x)}{g(x)}
By L'Hospital's Rule, we have:
So the answer is in the negative.
Now we have freshman level math. MVT goes first:
Question: Denote f(x) as a twice-differentiable function at domain [a,b] and f(a)=f(b)=0. Prove that there exists a point c ∈ (a,b) such that:
|f"(c)| ≥ \frac{8}{(b-a)^2} \max_{a\le x\le b} |f(x)|
Solution:
If f(x)≡ 0, the conclusion is trivially true.
Now consider:
f(x_0) = \max_{a\le x\le b} |f(x)|
where x_0 ∈ (a,b), i.e. f(x) attains maximum when x = x_0.
Also , since f(x) is twice-differentiable, we have f'(x_0) = 0. Now we expand the second-degree Taylor series for f(x) at x = x_0 with Lagrange's form of remainder, we have:
f(x) = f(x_0) + f'(x_0) (x - x_0) + \frac{f"(d)}{2} (x-x_0)^2
where
d = x_0 + g(x-x_0) and 0<g<1.
Since f(a) = 0, put x = a into the above series, we have:
0 = f(a) = f(x_0) + f'(x_0) (a - x_0) + \frac{f"(h)}{2} (a-x_0)^2
where a < h < x_0.
Now,
|f"(h)| = \frac {2|f(x_0)|}{(x_0 - a)^2}
Likewise:
|f"(j)| = \frac {2|f(x_0)|}{(b - x_0)^2}
where x_0 <j < b.
Then for:
|f"(d)| = \max_{a\le x\le b} |f"(x)|
we apply AM-GM inequalities such that:
|f"(d)| ≥ \frac{|f"(h)| + |f"(j)|}{2}
= |f(x_0)| [\frac{1}{(x_0-a)^2}+ \frac{1}{(b - x_0)^2}]
≥ |f(x_0)| ‧ \frac{8}{(b-a)^2} = \frac{8|f(x_0)|}{(b-a)^2}
where the equality holds if and only if x_0 = \frac{a+b}{2}.
Q.E.D.
Now try one more:
Question: Let f(x) be a twice-differentiable function at domain [0,1] satisfying |f(x)|≤ a and |f"(x)|≤ b where both a and b are non-negative constants. Denote c as any point within domain (0,1), prove that:
|f'(c)| ≤ 2a +\frac {b}{2}
Solution:
Using the same trick from the previous question, we have:
f(x) = f(c) + f'(c)(x-c) + \frac{f"(d)}{2!}(x-c)^2
where d = c + g(x-c), 0 < g < 1.
(Recall what a position vector looks like)
Put x=0, we have:
f(0) = f(c) + f'(c)(0-c) + \frac{f"(h)}{2!}(0-c)^2
where 0<h<c<1
Put x=1, we have:
f(1) = f(c) + f'(c)(1-c) + \frac{f"(j)}{2!}(1-c)^2
where 0<c<j<1
Subtracting each other, we have:
f(1) - f(0) = f'(c) + \frac{1}{2!} [f"(j)(1-c)^2] - f"(h)c^2]
Then by triangle inequalities:
|f'(c)| = |f(1) - f(0) - \frac{1}{2!} [f"(j)(1-c)^2] - f"(h)c^2]|
≤ |f(1)| + |f(0)| + |\frac{1}{2!}f"(j)| (1-c)^2 + |\frac{1}{2!}f"(h)| c^2
≤ 2a + \frac{b}{2} [(1-c)^2 + c^2]
Note that for 0<c<1, (1-c)^2 + c^2≤ 1
Q.E.D.
Now try 5 more questions on mean value theorem for integral calculus:
Answers in English will be announced later.
Solution to MVT(IC) Question 1:
Since f(x) is differentiable on interval [a,b], f(x) is also continuous on interval [a,b], then F(x) = f(x) cosx is continuous on interval [a, \frac{a+b}{2} ]. By MVT(IC), there exists a point c ∈ [a, \frac{a+b}{2} ] such that:
F(b) = f(b) cos b = \frac{1}{\frac{a+b}{2}-a} \int_{a}^{\frac{a+b}{2}} f(x) cosx dx
= \frac{2}{b-a} \int_{a}^{\frac{a+b}{2}} f(x) cosx dx
= \frac{2}{b-a} F(c) (\frac{a+b}{2}-a) = F(c).
On interval [c.b], by Rolle's Theorem, there exists at least a point d ∈ (c,b) ⊂ (a,b) such that:
F'(d) = f'(d)cosd + f(d)cosd'
=f'(d)cosd - f(d)sind = 0
Rearranging gives:
f'(d) = f(d)tand,
d ∈ (a,b)
Q.E.D.
Solution to MVT(IC) Question 2:
Part (I):
The question is equivalent to proving:
x_0 f(x_0) = \int_{x_0}^{1} f(x)dx.
Now let:
g(x) = -x \int_{x}^{1} f(t)dt.
Then g(x) is continuous on domain [0,1] and differentiable on interval (0,1). Also, since g(0) = g(1) = 0, by Rolle's Theorem again, there exists x_0 ∈ (0,1) such that:
g(x_0) = x_0 f(x_0) - \int_{x_0}^{1} f(x)dx = 0
i.e.
x_0 f(x_0) = \int_{x_0}^{1} f(x)dx
The above proof uses the method of auxiliary function.
Part (II):
Denote:
F(x) = xf(x) - \int_{x}^{1} f(x)dx
Given:
f'(x) > -\frac{2f(x)}{x}
Then:
F'(x) = [xf'(x) + f(x)] - [0 - f(x)] = 2f(x) + xf'(x) >0
Then F(x) is strictly increasing on domain (0,1).
So the value of x_0 is unique.
Solution to MVT(IC) Question 3:
Denote
F(x) = \int_{0}^{x} f(t) dt
where 0\le x \le \pi.
Then F(0) = 0 and F(\pi) = 0.
Also, since:
0 = \int_{0}^{\pi} f(x)cosx dx = \int_{0}^{\pi} cosx dF(x) = \int_{0}^{\pi} F(x)sinxdx ...(*)
So there exist c ∈ (0, \pi) such that F(c)sinc = 0;
otherwise, in the domain (0, \pi),
F(x)sinx > 0 or F(x)sinx < 0, which contradicts the fact (*).
Note that for c ∈ (0, \pi) , sinc \neq 0, so F(c) = 0.
Then for F(x), for domains [0,c] and [c, \pi], applying Rolle's Theorem, there exists at least one point m ∈ (0, c) and one point n ∈ (c, \pi) such that F'(m) = F'(n) = 0, i.e. f(m) = f(n) = 0.
Done.
Solution to MVT(IC) Question 4:
Using integration by part:
\int_{0}^{a} f(x)dx = \int_{0}^{a} f(x)d(x-a)
= [(x-a)f(x)] |_{0}^{a} - \int_{0}^{a} (x-a)f'(x)dx
= af(0) - \int_{0}^{a} (x-a)f'(x)dx
Since f'(x) is continuous. it has a minimum m and maximum M on domain [0,a] such that:
m(a-x) \le (a-x)f'(x) \le M(a-x)
So:
\frac{ma^2}{2} \le \int_{0}^{a} (x-a)f'(x)dx \le \frac{Ma^2}{2}
Also, by Intermediate Value Theorem, there exists at least one point c ∈ [0,a] such that:
\int_{0}^{a} (x-a)f'(x)dx = \frac{a^2}{2}f'(c)
Hence,
\int_{0}^{a} f(x)dx = af(0) + \frac{a^2}{2}f'(c)
Solution to MVT(IC) Question 5:
Denote:
\max_{[a,b]} {|f(x)|} = |f(x_0)|
Then we are going to prove that:
(b-a)|f(x_0)| \le |\int_{a}^{b} f(x)dx| + (b-a) \int_{a}^{b} |f'(x)|dx
i.e.
|\int_{a}^{b} f(x_0)dx| - |\int_{a}^{b} f(x)dx| \le (b-a) \int_{a}^{b} |f'(x)|dx
In fact, by triangle inequalities:
|\int_{a}^{b} f(x_0)dx| - |\int_{a}^{b} f(x)dx| \le | \int_{a}^{b} [f(x_0) - f(x)]dx|
= |\int_{a}^{b} [\int_{x}^{x_0}f'(t)dt]dx| \le \int_{a}^{b} [\int_{a}^{b}|f'(t)|dt]dx
=(b-a) \int_{a}^{b}|f'(x)|dx
Q.E.D.
Solution to MVT(IC) Question 6:
Part (I):
Using the definition of an derivative:
F'(x) = \lim_{\Delta x\to 0} \frac{F(x + \Delta x) - F(x)}{\Delta x}
= \lim_{\Delta x\to 0} \frac{\int_{0}^{x + \Delta x} f(t)dt - \int_{0}^{x} f(t)dt}{\Delta x}
= \lim_{\Delta x\to 0} \frac{\int_{x}^{x + \Delta x} f(t)dt}{\Delta x}
Since f(x) is continuous, by Intermediate Value Theorem for Integral Calculus, there exists c ∊ (x, x + \Delta x) such that:
\int_{x}^{x + \Delta x} f(t)dt = f(c) ‧ \Delta x
So:
F'(x) = \lim_{\Delta x\to 0} \frac{\int_{x}^{x + \Delta x} f(t)dt}{\Delta x} = \lim_{\Delta x\to 0} \frac{f(c) ‧ \Delta x}{\Delta x} = \lim_{\Delta x\to 0} f(c) = f(x)
Part (II):
G'(x+2) = [2 \int_{0}^{x+2} f(t)dt - (x+2) \int_{0}^{2}f(t)dt]'
= 2f(x+2) - \int_{0}^{2}f(t)dt
G'(x) = [2 \int_{0}^{x} f(t)dt - x\int_{0}^{2}f(t)dt]'
= 2f(x) - \int_{0}^{2}f(t)dt
For periodic function, f(x+2) = f(x) such that:
G'(x+2) = G'(x)
Integration gives:
G(x+2) - G(x) = C
Put x = 0, C = G(0+2) - G(0) = 0
So G(x+2) = G(x).