# Evaluate the integral: $\rm \int_1^2 \log x \ dx$.

14 views
in Calculus
closed
Evaluate the integral: $\rm \int_1^2 \log x \ dx$.
1. 2 log 2 - 1
2. 2 log 2 + 1
3. 2 log 2 - 3
4. None of these.

by (54.3k points)
selected

Correct Answer - Option 1 : 2 log 2 - 1

Concept:

Integration by Parts:

∫ f(x) g(x) dx = f(x) ∫ g(x) dx - ∫ [f'(x) ∫ g(x) dx] dx.

Definite Integral:

If ∫ f(x) dx = g(x) + C, then $\rm ∫_a^b f(x)\ dx = [ g(x)]_a^b$ = g(b) - g(a).

Calculation:

Let's first integrate the expression under integral by parts.

I = ∫ log x dx =  ∫ (1)(log x) dx

Considering log x as the first function and 1 as the second function, we get:

= (log x) ∫ 1 dx - ∫ [$\rm\frac1x$ ∫ 1 dx] dx

= (log x) x - x + C

Putting the limits of the definite integral, we get:

$\rm ∫_1^2 \log x \ dx$

$\rm \left[x(\log x)-x\right]_1^2$

= (2 log 2 - 2) - (0 - 1)

= 2 log 2 - 1.