91Ó°ÊÓ

We have to evaluate the line integrals in given exercise.

${∫}_{C}f(x,y,z)ds,\mathrm{where}f(x,y,z)=(x+z)3^y$ andC is the curve parameterized by $x=2−t,y=4t,z=t+5,\mathrm{for}t=1\mathrm{to}t=4$.

Here we have

$$\begin{gathered} f\left(x\right(t),y(t),z(t\left)\right)=(2-t+t+5)3^{4t} \\ f\left(x\right(t),y(t),z(t\left)\right)=(2+5)3^{4t} \\ f\left(x\right(t),y(t),z(t\left)\right)={73}^{4t} \end{gathered}$$

and

$$\begin{gathered} \frac{ds}{d\mathrm{t}}=\frac{d}{d\mathrm{t}}(2-t,4t,t+5) \\ \frac{ds}{d\mathrm{t}}=\left(\frac{d}{d\mathrm{t}}(2-t),\frac{d}{d\mathrm{t}}4t,\frac{d}{d\mathrm{t}}(t+5)\right) \\ \frac{ds}{d\mathrm{t}}=\left(-1,4,1\right) \end{gathered}$$

$$\begin{gathered} {∫}_{C}f(x,y,z)ds=7{∫}_1^4{3}^{4t}(-1,4,1)\mathrm{d}t \\ {∫}_{C}f(x,y,z)ds=7\left(-{∫}_1^4{3}^{4t}\mathrm{d}t+4{∫}_1^4{3}^{4t}\mathrm{d}t+{∫}_1^4{3}^{4t}\mathrm{d}t\right) \\ let4t=u \\ 4dt=du \\ dt=\frac{1}{4}du \\ {∫}_{C}f(x,y,z)ds=7\left(-\frac{1}{4}{∫}_1^4{3}^u\mathrm{du}+\frac{4}{4}{∫}_1^4{3}^u\mathrm{du}+\frac{1}{4}{∫}_1^4{3}^u\mathrm{du}\right) \\ {∫}_{C}f(x,y,z)ds=7{\left(-\frac{1}{4}\left[\frac{3^u}{\log 3}\right]+1\left[\frac{3^u}{\log 3}\right]+\frac{1}{4}\left[\frac{3^u}{\log 3}\right]\right)}_1^4 \\ {∫}_{C}f(x,y,z)ds=7{\left[\frac{3^u}{\log 3}\right]}_1^4 \\ {∫}_{C}f(x,y,z)ds=7{\left[\frac{3^{4t}}{\log 3}\right]}_1^4 \\ {∫}_{C}f(x,y,z)ds=7\left[\frac{3^{4·4}}{\log 3}-\frac{3^{4·1}}{\log 3}\right] \\ {∫}_{C}f(x,y,z)ds=7·\frac{3^4}{\log 3}\left[3^4-1\right] \\ {∫}_{C}f(x,y,z)ds=7·\frac{3^4}{\log 3}\left[81-1\right] \\ {∫}_{C}f(x,y,z)ds=7·\frac{81}{\log 3}·80 \\ {∫}_{C}f(x,y,z)ds≈\frac{45,360}{0.48} \\ {∫}_{C}f(x,y,z)ds≈94,500 \end{gathered}$$