Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.