Expert answers to "Evaluate the integrals ∫01[tet2+e−tj+k]dt"

nagasenaz

Answered question

2021-02-08

Evaluate the integrals
$\int_{0}^{1} [t e^{t 2} + e^{- t j} + k] d t$

Answer & Explanation

timbalemX

Skilled2021-02-09Added 108 answers

Step 1
Given that,
$\int_{0}^{1} [t e^{t^{2}} i + e^{- t} \cdot j + k] d t$
Evaluate an integral vector function by evaluating the integration of each component of a vector.
$\int_{0}^{1} [t e^{t^{2}} i + e^{- t} \cdot j + k] d t$
$= \int_{0}^{1} t e^{t^{2}} d t + \int_{0}^{1} e^{- t} d t + \int_{0}^{1} 1 d t k$
That is, we have to find the following integrals:
$\int_{0}^{1} t e^{t^{2}} d t$
$\int_{0}^{1} e^{- t} d t$
$\int_{0}^{1} 1 d t$
Step 2
Now,
$\int_{0}^{1} 1 d t = {(t)}_{0}^{1} = (1 - 0) = 1$
$\int_{0}^{1} e^{- t} d t$
$= {(- e^{- t})}_{0}^{1}$
$= (- e^{- 1} - (- e^{- 0}))$
$= - e^{- 1} + 1$
$= 1 - \frac{1}{e}$
$\int_{0}^{1} t e^{t 2} d t$
Plug $u = t^{2}$
if t = 0 then u = 0
if t = 1 then u = 1
du=2tdt
$\int_{0}^{1} \frac{1}{2} e^{u} d u$
$= \frac{1}{2} \int_{0}^{1} e^{u} d u$
$= \frac{1}{2} {(e^{u})}_{0}^{1}$
$= \frac{1}{2} (e^{1} - e^{0})$
$= \frac{1}{2} (e - 1)$
Step 3
Therefore,
$\int_{0}^{1} t e^{t 2} d t i + \int_{0}^{2} e^{- t} d t j + \int_{0}^{1} 1 d t k$
$= (\frac{e - 1}{2}) i + (\frac{e - 1}{e}) j + (1) k$
Thus,
$\int_{0}^{1} (t e^{t 2} i + e^{- t j} + k) d t = ((e - 1) / 2) i + ((e - 1) / e) j + k$

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