Chapter 11: Q. 42 (page 872)

Evaluate the integrals:
$\int_{1}^{2} <t e^{t}, I n t, \tan^{- 1} t> d t$

Short Answer

Expert verified

$[2 e^{2} - e^{2}] i + [2 \ln 2 - 2] j + [2 \tan^{- 1} 2 - \ln \sqrt{5}] - j + (\frac{π}{4} - \ln \sqrt{2}) k$

Step by step solution

Given Information

Consider $\int_{1}^{2} <t e^{t}, \ln t, \tan^{- 1} t> d t$

The objective is to evaluate the definite integral from $1$ and $2$ .

The integral of a vector function:

$\int r (t) d t = \int x (t) i d t + y (t) j d t + z (t) k d t = i \int x (t) i d t + j \int y (t) d t + k \int z (t) d t + c$

Where $c$ is a vector constant and has the form $c = (c_{1}, c_{2}, c_{3})$ for scalar constants $c_{1,} c_{2}$ and $c_{3}$ .

Now, $\int <t e^{t}, \ln t, \tan^{- 1} t> d t = i \int t e^{t} d t + j \int i n t d t + k \int \tan^{- 1} t d t = i [t \int e^{t} d t - \int (\frac{d}{d t} (t)) \int e^{t} d t] + j [t \ln t - t] + k [t \tan^{- 1} t - \log \sqrt{1 + t^{2}}] = i [t e^{t} - e^{t}] + j [t \ln t - t] + k [t \tan^{- 1} t - \ln \sqrt{1 + t^{2}}]$

Expression

Now,

$\int_{1}^{2} <t e^{t}, \ln t, \tan^{- 1} t> d t = i [t e^{t} - e^{t}] + j [t \ln t - t] + k {[t \tan^{- 1} t - \ln \sqrt{1 + t^{2}}]}_{1}^{2} = [2 e^{2} - e^{2}] i + [2 \ln 2 - 2] j + [2 \tan^{- 1} 2 - \ln \sqrt{5}] - j + (\frac{π}{4} - \ln \sqrt{2}) k$

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Evaluate the integrals:
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Short Answer

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Evaluate the integrals:
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