Evaluate the integral.
-$$\int _ { 0 } ^ { 3 } \left[ \left( \frac { 3 } { \sqrt { 1 + t } } \right) i - \left( 5 t ^ { 2 } \right) j + \left( \frac { 12 t } { \left( 1 + t ^ { 2 } \right) ^ { 2 } } \right) k \right] d t$$
A) $6 \mathbf { i } - 45 \mathrm { j } + \frac { 9 } { 10 } \mathrm { k }$
B) $3 \mathbf { i } - 45 \mathrm { j } + \frac { 27 } { 5 } \mathbf { k }$
C) $6 \mathbf { i } - 45 \mathbf { j } + \frac { 27 } { 5 } \mathbf { k }$
D) $6 \mathbf { i } + 45 \mathbf { j } + \frac { 9 } { 10 } \mathbf { k }$

Question

Quizplus · Accepted Answer

To solve the integral, we evaluate each component separately:- For the $i$ component: $\int_0^3 \frac{3}{\sqrt{1+t}} dt$, let $u = 1 + t$, then $du = dt$, and when $t = 0$, $u = 1$, and when $t = 3$, $u = 4$. So, the integral becomes $\int_1^4 \frac{3}{\sqrt{u}} du = 6\sqrt{u}\Big|_1^4 = 6(\sqrt{4} - \sqrt{1}) = 6(2 - 1) = 6$.- For the $j$ component: $\int_0^3 5t^2 dt = \frac{5}{3}t^3\Big|_0^3 = \frac{5}{3}(3^3) = 45$.- For the $k$ component: $\int_0^3 \frac{12t}{(1+t^2)^2} dt$, let $u = 1 + t^2$, then $du = 2t dt$, and the integral becomes $\frac{1}{2}\int \frac{12}{u^2} du = -\frac{6}{u}\Big|_1^{10} = -6\left(\frac{1}{10} - 1\right) = \frac{54}{10} = \frac{27}{5}$.Therefore, the integral evaluates to $6 \mathbf { i } - 45 \mathbf { j } + \frac { 27 } { 5 } \mathbf { k }$.

Evaluate the Integral A) 6i−45j+910k6 \mathbf { i } - 45 \mathrm { j } + \frac { 9 } { 10 } \mathrm { k }6i−45j+109​k

Evaluate the Integral A) $6 \mathbf { i } - 45 \mathrm { j } + \frac { 9 } { 10 } \mathrm { k }$