# For problems 1 the area of the region below the parameric curve given

For problems 1 the area of the region below the parameric curve given by the set of parametric equations. For each problem you may assume that each curve traces out exactly once from right to left for the given range of t. For these problems you should only use the given parametric equations to determine the answer.
1) $$\displaystyle{x}={t}^{{{2}}}+{5}{t}-{1}$$
$$\displaystyle{y}={40}-{t}^{{{2}}}$$
$$\displaystyle-{2}\le{t}\le{5}$$

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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
$$\displaystyle{A}={\int_{{{a}}}^{{{b}}}}{g{{\left({t}\right)}}}{f}'{\left({t}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({40}-{t}^{{{2}}}\right)}{\left({2}{t}+{5}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{-{2}}}^{{{5}}}}{\left({80}{t}-{2}{t}^{{{2}}}-{5}{t}^{{{2}}}+{200}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={{\left[{80}{\frac{{{t}^{{{2}}}}}{{{2}}}}-{2}{\frac{{{t}^{{{4}}}}}{{{4}}}}-{5}{\frac{{{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}$$
$$\displaystyle={{\left[{40}{t}^{{{2}}}-{\frac{{{t}^{{{4}}}}}{{{2}}}}-{\frac{{{5}{t}^{{{3}}}}}{{{3}}}}+{200}{t}\right]}_{{-{2}}}^{{{5}}}}$$
$$\displaystyle={\left[{1000}-{\frac{{{625}}}{{{2}}}}-{\frac{{{625}}}{{{3}}}}+{1000}\right]}-{\left[{160}-{8}+{\frac{{{40}}}{{{3}}}}-{400}\right]}$$
$$\displaystyle={2000}-{\frac{{{3125}}}{{{6}}}}+{248}-{\frac{{{40}}}{{{3}}}}$$
$$\displaystyle={2248}+{534.166}$$
$$\displaystyle={2782.162}$$