Answer :
To evaluate the integral
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt, \][/tex]
we need to integrate each component of the vector separately.
1. Integral of the i-component:
[tex]\[ \int \frac{8}{1+t^2} dt \][/tex]
The integral of [tex]\(\frac{8}{1+t^2}\)[/tex] is [tex]\(8 \arctan(t)\)[/tex]. Thus, the i-component integral is:
[tex]\[ 8 \arctan(t) + C_1, \][/tex]
where [tex]\(C_1\)[/tex] is the constant of integration for the i-component.
2. Integral of the j-component:
[tex]\[ \int t e^{t^2} dt \][/tex]
Let [tex]\(u = t^2\)[/tex]. Then [tex]\(du = 2t \, dt\)[/tex] or [tex]\(dt = \frac{du}{2t}\)[/tex]. Substituting in the integral, we have:
[tex]\[ \int t e^{t^2} dt = \frac{1}{2} \int e^u du \][/tex]
The integral of [tex]\(e^u\)[/tex] is [tex]\(e^u\)[/tex]. Substituting back for [tex]\(u\)[/tex], we get:
[tex]\[ \frac{1}{2} e^{t^2} + C_2, \][/tex]
where [tex]\(C_2\)[/tex] is the constant of integration for the j-component.
3. Integral of the k-component:
[tex]\[ \int 5 \sqrt{t} dt \][/tex]
Recall that [tex]\(\sqrt{t} = t^{1/2}\)[/tex], so the integral becomes:
[tex]\[ \int 5 t^{1/2} dt \][/tex]
Using the power rule for integration, [tex]\(\int t^n dt = \frac{t^{n+1}}{n+1}\)[/tex], we get:
[tex]\[ 5 \cdot \frac{t^{3/2}}{3/2} = \frac{10}{3} t^{3/2} \][/tex]
Thus, the k-component integral is:
[tex]\[ \frac{10}{3} t^{3/2} + C_3, \][/tex]
where [tex]\(C_3\)[/tex] is the constant of integration for the k-component.
Combining all components, the integral is:
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt = \left( 8 \arctan(t) + C_1 \right)i + \left( \frac{1}{2} e^{t^2} + C_2 \right)j + \left( \frac{10}{3} t^{3/2} + C_3 \right)k, \][/tex]
where [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex] are constants of integration.
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt, \][/tex]
we need to integrate each component of the vector separately.
1. Integral of the i-component:
[tex]\[ \int \frac{8}{1+t^2} dt \][/tex]
The integral of [tex]\(\frac{8}{1+t^2}\)[/tex] is [tex]\(8 \arctan(t)\)[/tex]. Thus, the i-component integral is:
[tex]\[ 8 \arctan(t) + C_1, \][/tex]
where [tex]\(C_1\)[/tex] is the constant of integration for the i-component.
2. Integral of the j-component:
[tex]\[ \int t e^{t^2} dt \][/tex]
Let [tex]\(u = t^2\)[/tex]. Then [tex]\(du = 2t \, dt\)[/tex] or [tex]\(dt = \frac{du}{2t}\)[/tex]. Substituting in the integral, we have:
[tex]\[ \int t e^{t^2} dt = \frac{1}{2} \int e^u du \][/tex]
The integral of [tex]\(e^u\)[/tex] is [tex]\(e^u\)[/tex]. Substituting back for [tex]\(u\)[/tex], we get:
[tex]\[ \frac{1}{2} e^{t^2} + C_2, \][/tex]
where [tex]\(C_2\)[/tex] is the constant of integration for the j-component.
3. Integral of the k-component:
[tex]\[ \int 5 \sqrt{t} dt \][/tex]
Recall that [tex]\(\sqrt{t} = t^{1/2}\)[/tex], so the integral becomes:
[tex]\[ \int 5 t^{1/2} dt \][/tex]
Using the power rule for integration, [tex]\(\int t^n dt = \frac{t^{n+1}}{n+1}\)[/tex], we get:
[tex]\[ 5 \cdot \frac{t^{3/2}}{3/2} = \frac{10}{3} t^{3/2} \][/tex]
Thus, the k-component integral is:
[tex]\[ \frac{10}{3} t^{3/2} + C_3, \][/tex]
where [tex]\(C_3\)[/tex] is the constant of integration for the k-component.
Combining all components, the integral is:
[tex]\[ \int \left( \frac{8}{1+t^2} i + t e^{t^2} j + 5 \sqrt{t} k \right) dt = \left( 8 \arctan(t) + C_1 \right)i + \left( \frac{1}{2} e^{t^2} + C_2 \right)j + \left( \frac{10}{3} t^{3/2} + C_3 \right)k, \][/tex]
where [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex] are constants of integration.
