Step 1

Given,

As per our honor code, we are answering only the first question.

The parametric equations are \(\displaystyle{x}={t}^{{{2}}}+{5}{t}-{1},\ {y}={40}-{t}^{{{2}}},\ -{2}\le{t}\le{5}\) We have to find the area of the region bounded by these parametric equations.

Step 2

If the parametric equations are \(\displaystyle{x}={f{{\left({t}\right)}}},\ {y}={g{{\left({t}\right)}}}\) and \(\displaystyle{a}\le{t}\le{b}\) then the area of the bounded region is given by

\(\displaystyle{A}={\int_{{{a}}}^{{{b}}}}{g{{\left({t}\right)}}}{f}'{\left({t}\right)}{\left.{d}{t}\right.}\)

Step 3

\(\displaystyle{f{{\left({t}\right)}}}={x}={t}^{{{2}}}={5}{t}-{1}\Rightarrow{f}'{\left({t}\right)}={2}{t}+{5}\)

\(\displaystyle{g{{\left({t}\right)}}}={y}={40}-{t}^{{{2}}}\)

Hence the area is

\(\displaystyle{A}={\int_{{{a}}}^{{{b}}}}{g{{\left({t}\right)}}}{f}'{\left({t}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({40}-{t}^{{{2}}}\right)}{\left({2}{t}+{5}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({80}{t}-{2}{t}^{{{2}}}-{5}{t}^{{{2}}}+{200}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={{\left[{80}{\frac{{{t}^{{{2}}}}}{{{2}}}}-{2}{\frac{{{t}^{{{4}}}}}{{{4}}}}-{5}{\frac{{{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}\)

\(\displaystyle={{\left[{40}{t}^{{{2}}}-{\frac{{{t}^{{{4}}}}}{{{2}}}}-{\frac{{{5}{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}\)

\(\displaystyle={\left[{1000}-{\frac{{{625}}}{{{2}}}}-{\frac{{{625}}}{{{3}}}}+{1000}\right]}-{\left[{160}-{8}+{\frac{{{40}}}{{{3}}}}-{400}\right]}\)

\(\displaystyle={2000}-{\frac{{{3125}}}{{{6}}}}+{248}-{\frac{{{40}}}{{{3}}}}\)

\(\displaystyle={2248}+{534.166}\)

\(\displaystyle={2782.162}\)

Given,

As per our honor code, we are answering only the first question.

The parametric equations are \(\displaystyle{x}={t}^{{{2}}}+{5}{t}-{1},\ {y}={40}-{t}^{{{2}}},\ -{2}\le{t}\le{5}\) We have to find the area of the region bounded by these parametric equations.

Step 2

If the parametric equations are \(\displaystyle{x}={f{{\left({t}\right)}}},\ {y}={g{{\left({t}\right)}}}\) and \(\displaystyle{a}\le{t}\le{b}\) then the area of the bounded region is given by

\(\displaystyle{A}={\int_{{{a}}}^{{{b}}}}{g{{\left({t}\right)}}}{f}'{\left({t}\right)}{\left.{d}{t}\right.}\)

Step 3

\(\displaystyle{f{{\left({t}\right)}}}={x}={t}^{{{2}}}={5}{t}-{1}\Rightarrow{f}'{\left({t}\right)}={2}{t}+{5}\)

\(\displaystyle{g{{\left({t}\right)}}}={y}={40}-{t}^{{{2}}}\)

Hence the area is

\(\displaystyle{A}={\int_{{{a}}}^{{{b}}}}{g{{\left({t}\right)}}}{f}'{\left({t}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({40}-{t}^{{{2}}}\right)}{\left({2}{t}+{5}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({80}{t}-{2}{t}^{{{2}}}-{5}{t}^{{{2}}}+{200}\right)}{\left.{d}{t}\right.}\)

\(\displaystyle={{\left[{80}{\frac{{{t}^{{{2}}}}}{{{2}}}}-{2}{\frac{{{t}^{{{4}}}}}{{{4}}}}-{5}{\frac{{{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}\)

\(\displaystyle={{\left[{40}{t}^{{{2}}}-{\frac{{{t}^{{{4}}}}}{{{2}}}}-{\frac{{{5}{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}\)

\(\displaystyle={\left[{1000}-{\frac{{{625}}}{{{2}}}}-{\frac{{{625}}}{{{3}}}}+{1000}\right]}-{\left[{160}-{8}+{\frac{{{40}}}{{{3}}}}-{400}\right]}\)

\(\displaystyle={2000}-{\frac{{{3125}}}{{{6}}}}+{248}-{\frac{{{40}}}{{{3}}}}\)

\(\displaystyle={2248}+{534.166}\)

\(\displaystyle={2782.162}\)