Definite IntegrationeasyFree

Definite Integration: Let Positive Real Numbers

JEE Maths question with a full step-by-step solution.

Question

Let

a

and

b

be positive real numbers and

\displaystyle f\left(a,b\right) = \int_a^b\frac{e^{\frac xa}-e^{\frac bx}}{x}\,dx

, then

f\left(1,2\right) = 0

correct

f\left(3,1\right) = 2

f\left(2,1\right) = 0

correct

f\left(1,3\right) = 2

Solution

Step 1: The two exponents are

\dfrac xa

and

\dfrac bx

. The substitution

t = \frac{ab}{x} \quad\Longleftrightarrow\quad x = \frac{ab}{t}

turns

\dfrac xa

into

\dfrac bt

and

\dfrac bx

into

\dfrac ta

, the same two expressions with the roles of the exponentials exchanged. Step 2:

dx = -\frac{ab}{t^2}\,dt ,\qquad \frac{1}{x} = \frac{t}{ab}

and the limits swap:

x = a \Rightarrow t = b

x = b \Rightarrow t = a

I = \int_{t=b}^{t=a}\left(e^{\frac bt}-e^{\frac ta}\right)\cdot\frac{t}{ab}\cdot\left(-\frac{ab}{t^2}\right)dt = \int_b^a\left(-\frac{e^{\frac bt}-e^{\frac ta}}{t}\right)dt

Step 3: Interchanging the limits,

I = \int_a^b\frac{e^{\frac bt}-e^{\frac ta}}{t}\,dt = -\int_a^b\frac{e^{\frac ta}-e^{\frac bt}}{t}\,dt = -I

Step 4:

a,b>0

, so the integrand is continuous on the closed interval joining

a

and

b

and

I

is a finite number. Hence

2I = 0 \quad\Longrightarrow\quad I = 0

for **every** pair of positive reals

a,b

; the integral never depends on them. Step 5:

f\left(1,2\right) = 0 ,\qquad f\left(2,1\right) = 0

f\left(3,1\right) = 0 \ne 2 ,\qquad f\left(1,3\right) = 0 \ne 2

Numerically,

f\left(1,2\right)

f\left(3,1\right)

f\left(0.7,5.3\right)

all evaluate to zero to Answer: (1) and (3)

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