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Sagot :
To evaluate the integral of the given vector function, we will integrate each component separately with respect to [tex]\( t \)[/tex]. Let's denote the integral as follows:
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt \][/tex]
### Step-by-Step Solution
1. Integrating the [tex]\( i \)[/tex]-component:
[tex]\[ \int \frac{8}{1+t^2} \, dt \][/tex]
The integrand [tex]\(\frac{8}{1+t^2}\)[/tex] is a well-known form which integrates to [tex]\(8 \arctan(t)\)[/tex]. Hence,
[tex]\[ \int \frac{8}{1+t^2} \, dt = 8 \arctan(t) \][/tex]
2. Integrating the [tex]\( j \)[/tex]-component:
[tex]\[ \int t e^{t^2} \, dt \][/tex]
To integrate [tex]\( t e^{t^2} \)[/tex], we can use the substitution [tex]\( u = t^2 \)[/tex]. Therefore, [tex]\( du = 2t \, dt \)[/tex] or [tex]\( \frac{du}{2} = t \, dt \)[/tex]. Rewriting the integral in terms of [tex]\( u \)[/tex],
[tex]\[ \int t e^{t^2} \, dt = \int e^{u} \frac{du}{2} = \frac{1}{2} \int e^{u} \, du \][/tex]
The integral of [tex]\( e^{u} \)[/tex] is simply [tex]\( e^u \)[/tex]. Re-substituting [tex]\( u = t^2 \)[/tex],
[tex]\[ \frac{1}{2} e^u = \frac{1}{2} e^{t^2} \][/tex]
Thus,
[tex]\[ \int t e^{t^2} \, dt = \frac{1}{2} e^{t^2} \][/tex]
3. Integrating the [tex]\( k \)[/tex]-component:
[tex]\[ \int 5 \sqrt{t} \, dt \][/tex]
Recall that [tex]\(\sqrt{t} = t^{1/2} \)[/tex]. Hence,
[tex]\[ \int 5 t^{1/2} \, dt = 5 \int t^{1/2} \, dt \][/tex]
Using the power rule for integration [tex]\(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\)[/tex],
[tex]\[ 5 \int t^{1/2} \, dt = 5 \cdot \frac{t^{3/2}}{3/2} = \frac{10}{3} t^{3/2} \][/tex]
4. Combining the results:
Integrating each component separately, we get:
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt = 8 \arctan(t) i + \frac{1}{2} e^{t^2} j + \frac{10}{3} t^{3/2} k + C \][/tex]
Here, [tex]\( C \)[/tex] represents the constant of integration which is a vector-valued function. It can be written as [tex]\( C = C i + C j + C k \)[/tex].
Therefore, the final answer to the integral is:
[tex]\[ \boxed{8 \arctan(t) i + \frac{1}{2} e^{t^2} j + \frac{10}{3} t^{3/2} k + C} \][/tex]
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt \][/tex]
### Step-by-Step Solution
1. Integrating the [tex]\( i \)[/tex]-component:
[tex]\[ \int \frac{8}{1+t^2} \, dt \][/tex]
The integrand [tex]\(\frac{8}{1+t^2}\)[/tex] is a well-known form which integrates to [tex]\(8 \arctan(t)\)[/tex]. Hence,
[tex]\[ \int \frac{8}{1+t^2} \, dt = 8 \arctan(t) \][/tex]
2. Integrating the [tex]\( j \)[/tex]-component:
[tex]\[ \int t e^{t^2} \, dt \][/tex]
To integrate [tex]\( t e^{t^2} \)[/tex], we can use the substitution [tex]\( u = t^2 \)[/tex]. Therefore, [tex]\( du = 2t \, dt \)[/tex] or [tex]\( \frac{du}{2} = t \, dt \)[/tex]. Rewriting the integral in terms of [tex]\( u \)[/tex],
[tex]\[ \int t e^{t^2} \, dt = \int e^{u} \frac{du}{2} = \frac{1}{2} \int e^{u} \, du \][/tex]
The integral of [tex]\( e^{u} \)[/tex] is simply [tex]\( e^u \)[/tex]. Re-substituting [tex]\( u = t^2 \)[/tex],
[tex]\[ \frac{1}{2} e^u = \frac{1}{2} e^{t^2} \][/tex]
Thus,
[tex]\[ \int t e^{t^2} \, dt = \frac{1}{2} e^{t^2} \][/tex]
3. Integrating the [tex]\( k \)[/tex]-component:
[tex]\[ \int 5 \sqrt{t} \, dt \][/tex]
Recall that [tex]\(\sqrt{t} = t^{1/2} \)[/tex]. Hence,
[tex]\[ \int 5 t^{1/2} \, dt = 5 \int t^{1/2} \, dt \][/tex]
Using the power rule for integration [tex]\(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\)[/tex],
[tex]\[ 5 \int t^{1/2} \, dt = 5 \cdot \frac{t^{3/2}}{3/2} = \frac{10}{3} t^{3/2} \][/tex]
4. Combining the results:
Integrating each component separately, we get:
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt = 8 \arctan(t) i + \frac{1}{2} e^{t^2} j + \frac{10}{3} t^{3/2} k + C \][/tex]
Here, [tex]\( C \)[/tex] represents the constant of integration which is a vector-valued function. It can be written as [tex]\( C = C i + C j + C k \)[/tex].
Therefore, the final answer to the integral is:
[tex]\[ \boxed{8 \arctan(t) i + \frac{1}{2} e^{t^2} j + \frac{10}{3} t^{3/2} k + C} \][/tex]
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