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High School
Calculus and Analysis
Calculus 2
Applications of integrals

prsategazd

Answered question

2021-12-28

Evaluate the following integrals. Include absolute values only when needed.

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x

Answer & Explanation

Thomas Lynn

Beginner2021-12-29Added 28 answers

Step 1:To determine
Evaluate:

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x

Step 2:Calculation
Consider the given integral:

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x

Let

x + 1 = u \Rightarrow d x = d u

Also, when

x = 0 \Rightarrow u = 1

& when

x = 3 \Rightarrow u = 4

So, the integral becomes:

\int_{1}^{4} \frac{2 (u - 1) - 1}{u} d u

\Rightarrow \int_{1}^{4} \frac{2 u - 3}{u} d u

\Rightarrow \int_{1}^{4} \frac{2 u}{u} - \frac{3}{u} d u

\Rightarrow \int_{1}^{4} 2 - \frac{3}{u} d u

= 2 u - 3 \ln (u) ∣_{1}^{4}

= 2 (4) - 3 \ln (4) - 2 (1) + 3 \ln (1)

\Rightarrow 8 - 3 \ln (4) - 2 + 3 (0)

\Rightarrow 6 - 3 \ln (2^{2})

\Rightarrow 6 - 6 \ln (2)

Hence,

\int_{0}^{3} \frac{2 x - 1}{x + 1} d x = 6 - 6 \ln (2)

zurilomk4

Beginner2021-12-30Added 35 answers

\int \frac{2 x - 1}{x + 1} d x

Substitution

u = x + 1 \Rightarrow \frac{d u}{d x} = 1 \Rightarrow d x = d u

= \int \frac{2 u - 3}{u} d u

We use the distributive property:

= \int (2 - \frac{3}{u}) d u

Lets

karton

Expert2022-01-04Added 613 answers

$\begin{matrix} \int_{0}^{3} \frac{2 x - 1}{x + 1} d x \\ \int \frac{2 x - 1}{x + 1} d x \\ \int \frac{2 t - 3}{t} d t \\ \int \frac{2 t}{t} - \frac{3}{t} d t \\ \int 2 - \frac{3}{t} d t \\ \int 2 d t - \int \frac{3}{t} d t \\ 2 t - 3 \ln (| t |) \\ 2 (x + 1) - 3 \ln (| x + 1 |) \\ 2 x = 2 - 3 \ln (| x + 1 |) \\ 2 x + 2 - 3 \ln (| x + 1 |) \\ (2 x + 2 - 3 \ln (| x + 1 |)) |_{0}^{3} \\ 2 \times 3 + 2 - 3 \ln (| 3 + 1 |) - (2 \times 0 + 2 - 3 \ln (| 0 + 1 |)) \end{matrix}$
Simplify
$6 - 6 \ln (2)$

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