18.01 PROBLEM SET 5 – SOLUTIONS

2. Consider

∫_{a}^{b} f(x)dx.

This represents the area under the graph of f on the interval [a, b]. By changing

the variable we change to another area. The term \frac{du}{dx} in the formula corresponds

to the ratio of the two areas considered. It is needed as an “error-correcting” term. We have to change the limits, since we consider a different function on a different interval.

(2 points)

3.

(a) A = ∫_{1}^{b} ln xdx

B = ∫_{0}^{ln b} e^{y}dy

A + B = area of rectangle b by ln b = b · ln b.

(b)

B = ∫_{0}^{ln b} e^{y}dy = e^{y}|_{0}^{ln b} = e^{ln b} − e^{0} = b − 1.

Hence

∫_{1}^{b} ln xdx = b ln b − b + 1.

(c) By FTC2

\frac{d}{db} ∫_{1}^{b} ln xdx = ln b.

\frac{d}{db}(b ln b − b + 1) = b · \frac{1}{b} + ln b − 1 = ln b.

.

Therefore

∫_{1}^{b} ln xdx = b ln b − b + 1 + c

for some constant c.

Substitute b = 1, this gives ∫_{1}^{1} ln xdx = 0 and (b ln b−b+1)|_{b=1} = 1·ln 1−1+1 = 0. Hence c = 0.

Typeset by AMS-TEX

(d)

\frac{d}{db} ∫_{1}^{ln b} f(y)dy = f(ln b) · \frac{d}{db} ln b = f(ln b) · \frac{1}{b}.

Special case: f(y) = e^{y}.

∫_{1}^{ln b} f(y)dy = ∫_{1}^{ln b} e^{y}dy = b − 1,

so

\frac{d}{db} ∫_{1}^{ln b} f(y)dy = \frac{d}{db}b − 1 = 1.

On the other hand

f(ln b) · \frac{1}{b} = e^{ln b} · \frac{1}{b} == 1.

(8 points)

4. x = tan θ, then dx = sec^{2} θdθ and \sqrt{1 + x^{2}} = sec θ. Hence

∫ \frac{dx}{\sqrt{1 + x^{2}}} = ∫ \frac{sec^{2} θdθ}{sec θ} = ∫ sec θdθ
= ln(sec θ + tan θ) + c = ln(\sqrt{1 + x^{2}} + x) + c.

From problem set 2:

\frac{d}{dx} sinh^{−1} x = \frac{1}{\sqrt{1 + x^{2}}} ,

so

sinh^{−1} x = ln(\sqrt{1 + x^{2}} + x) + c

for some c.

Now sinh^{−1}(0) = 0, and ln(\sqrt{1 + x^{2}} + x)|_{x=0} = 0, so c = 0. Therefore

sinh^{−1} x = ln(\sqrt{1 + x^{2}} + x).

(5 points)