Question Statement

Evaluate the definite integral:

$\int_{1}^{3} ln x d x$

Background and Explanation

This problem requires integration by parts, a technique for integrating products of functions. Since $ln x$ does not have an elementary antiderivative on its own, we rewrite it as the product $ln x \cdot 1$ and apply the integration by parts formula.

Solution

To evaluate $\int_{1}^{3} ln x d x$ , we use integration by parts. The key insight is to treat $ln x$ as a product with the constant function $1$ :

$\int_{1}^{3} ln x d x = \int_{1}^{3} ln x \cdot 1 d x$

Setting up integration by parts:

Let $u = ln x$ and $v = 1$ (where $v$ is the second function to be integrated).

Recall the integration by parts formula for definite integrals: $\int_{a}^{b} u v d x = [u \cdot \int v d x - \int (\int v d x) \cdot \frac{d u}{d x} d x]_{a}^{b}$

Applying the formula:

$\int_{1}^{3} ln x d x = [ln x \cdot \int 1 d x - \int (\int 1 d x) \times \frac{d}{d x} (ln x) d x]_{1}^{3} = [ln x \cdot x - \int x \cdot \frac{1}{x} d x]_{1}^{3} = [x ln x - \int 1 d x]_{1}^{3} = [x ln x - x]_{1}^{3}$

Evaluating at the bounds:

Now we substitute the upper limit ( $x = 3$ ) and lower limit ( $x = 1$ ):

$= (3 ln 3 - 3) - (1 \cdot ln 1 - 1) = 3 ln 3 - 3 - (0 - 1) (since ln 1 = 0) = 3 ln 3 - 3 + 1 = 3 ln 3 - 2$

Therefore, the value of the integral is $3 ln 3 - 2$ .

Key Formulas or Methods Used

Integration by parts formula: $\int u d v = uv - \int v d u$ (or equivalently $\int u \cdot v d x = u \int v d x - \int (\int v d x) u^{'} d x$ )
Derivative of natural logarithm: $\frac{d}{d x} (ln x) = \frac{1}{x}$
Logarithm property: $ln 1 = 0$
Power rule: $\int 1 d x = x$

Summary of Steps

Rewrite the integrand as $ln x \cdot 1$ to prepare for integration by parts
Set $u = ln x$ (to differentiate) and $d v = d x$ (so $v = x$ )
Apply integration by parts: $\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x$
Simplify the remaining integral: $\int 1 d x = x$
Obtain the antiderivative: $x ln x - x$
Evaluate the definite integral using the Fundamental Theorem of Calculus: $[x ln x - x]_{1}^{3}$
Substitute bounds and simplify: $(3 ln 3 - 3) - (0 - 1) = 3 ln 3 - 2$

Solution

To evaluate

\int_{1}^{3} ln x d x

, we use integration by parts. The key insight is to treat

ln x

as a product with the constant function

1

\int_{1}^{3} ln x d x = \int_{1}^{3} ln x \cdot 1 d x

Setting up integration by parts:

Let

u = ln x

and

v = 1

(where

v

is the second function to be integrated).

Recall the integration by parts formula for definite integrals:

\int_{a}^{b} u v d x = [u \cdot \int v d x - \int (\int v d x) \cdot \frac{d u}{d x} d x]_{a}^{b}

Applying the formula:

\int_{1}^{3} ln x d x = [ln x \cdot \int 1 d x - \int (\int 1 d x) \times \frac{d}{d x} (ln x) d x]_{1}^{3} = [ln x \cdot x - \int x \cdot \frac{1}{x} d x]_{1}^{3} = [x ln x - \int 1 d x]_{1}^{3} = [x ln x - x]_{1}^{3}

Evaluating at the bounds:

Now we substitute the upper limit (

x = 3

) and lower limit (

x = 1

= (3 ln 3 - 3) - (1 \cdot ln 1 - 1) = 3 ln 3 - 3 - (0 - 1) (since ln 1 = 0) = 3 ln 3 - 3 + 1 = 3 ln 3 - 2

Therefore, the value of the integral is

3 ln 3 - 2

Summary of Steps

Rewrite the integrand as

ln x \cdot 1

to prepare for integration by parts

Set

u = ln x

(to differentiate) and

d v = d x

(so

v = x

)

Apply integration by parts:

\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x

Simplify the remaining integral:

\int 1 d x = x

Obtain the antiderivative:

x ln x - x

Evaluate the definite integral using the Fundamental Theorem of Calculus:

[x ln x - x]_{1}^{3}

Substitute bounds and simplify:

(3 ln 3 - 3) - (0 - 1) = 3 ln 3 - 2

3.7 Q-15

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

3.7 Q-15

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps