Question

Evaluate the integral:

∫(t to 0) (3si+(3s^2)j+17k)ds

Answer 1

$$\displaystyle\int_{0}^t(3s\vec{i}+3s^2\vec{j}+17\vec{k})ds=$$

$$=\displaystyle\int_{0}^t3sds\ \vec{i}+\displaystyle\int_{0}^t3s^2ds\ \vec{j}+\displaystyle\int_{0}^t17ds\ \vec{k}=$$

$$=[\dfrac{3s^2}{2}]\begin{matrix} t\\ 0 \end{matrix}\ \vec{i}+[\dfrac{3s^3}{3}]\begin{matrix} t\\ 0 \end{matrix}\ \vec{j}+[17s]\begin{matrix} t\\ 0 \end{matrix}\ \vec{k}=$$

$$=\dfrac{3t^2}{2}\ \vec{i}+t^3\ \vec{j}+17t\ \vec{k}$$

$$\displaystyle\int_{0}^t(3s\vec{i}+3s^2\vec{j}+17\vec{k})ds=\dfrac{3t^2}{2}\ \vec{i}+t^3\ \vec{j}+17t\ \vec{k}$$

$$\displaystyle\int_{t}^0(3s\vec{i}+3s^2\vec{j}+17\vec{k})ds=-\dfrac{3t^2}{2}\ \vec{i}-t^3\ \vec{j}-17t\ \vec{k}$$