 # Problem 1 (i) Determine the (x, y) coordinates of the points of intersection for the curves described by the following equations. wstecznyg5 2022-07-25 Answered
Problem 1
i) Determine the (x, y) coordinates of the points of intersection for the curves described by the following equations.
${y}_{1}={x}^{2}-4x+3\phantom{\rule{0ex}{0ex}}{y}_{2}={x}^{2}+2x+3$
ii) Set up a definite integral to give the area of the region bounded by the two curves
iii) Evaluate the definite integral to give the area
Proble 2
Same as i, ii, iii in problem 1 above for following equations:
${y}_{1}={x}^{2}\phantom{\rule{0ex}{0ex}}{y}_{2}={x}^{3}$
Problem 3
Same as i, ii, iii in problem 1 above for following equations:
$x=4-{y}^{2}\phantom{\rule{0ex}{0ex}}x=y-2$
Problem 4
Same as i, ii, iii in problem 3 above for the following equations:
$y={x}^{2}\phantom{\rule{0ex}{0ex}}y=6-x$
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Step 1

i)
ii) $f\left(x\right)-g\left(x\right)⇒-{x}^{2}+2x+\text{⧸}3-{x}^{2}+4x-\text{⧸}3\phantom{\rule{0ex}{0ex}}=-2{x}^{2}+6x$
iii) $A={\int }_{0}^{3}-2{x}^{2}+b×dx⇒-2{\int }_{0}^{3}{x}^{2}dx+6{\int }_{0}^{3}xdx\phantom{\rule{0ex}{0ex}}⇒2{|\frac{{x}^{3}}{3}|}_{0}^{3}+6{|\frac{{x}^{2}}{2}|}_{0}^{3}⇒\frac{-2}{3}\left[27\right]+\frac{6}{2}\left[9\right]\phantom{\rule{0ex}{0ex}}=\frac{-2}{\text{⧸}3}\left[\text{⧸}{27}^{9}\right]+3\left[9\right]\phantom{\rule{0ex}{0ex}}=-18+27⇒9{m}^{2}$
Step 2

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Step 1
Problem 3
When intersect

points of intersection are (-5, -3), (0, 2)
ii) are of the region $={\int }_{-3}^{2}\left(\left(4-{y}^{2}\right)-\left(y-2\right)\right)dy$
iii) are of the region:
$⇒={\int }_{-3}^{2}\left(4-{y}^{2}-y+2\right)dy\phantom{\rule{0ex}{0ex}}⇒={\int }_{-3}^{2}\left(6-{y}^{2}-y\right)dy\phantom{\rule{0ex}{0ex}}⇒={\int }_{-3}^{2}\left(6y-\left(\frac{1}{3}\right){y}^{3}-\left(\frac{1}{2}\right){y}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒=\left(6\left(2\right)-\left(\frac{1}{3}\right)\left(2{\right)}^{3}-\left(\frac{1}{2}\right)\left(2{\right)}^{2}\right)-\left(6\left(3\right)-\left(\frac{1}{3}\right)\left(-3{\right)}^{3}-\left(\frac{1}{2}\right)\left(-3{\right)}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒=\left(12-\left(\frac{8}{3}\right)-2\right)-\left(-18+9-\left(\frac{9}{2}\right)\right)\phantom{\rule{0ex}{0ex}}⇒=12-\left(\frac{8}{3}\right)-2+18-9+\left(\frac{9}{2}\right)\phantom{\rule{0ex}{0ex}}⇒=19-\left(\frac{8}{3}\right)+\left(\frac{9}{2}\right)\phantom{\rule{0ex}{0ex}}⇒=\frac{125}{6}\text{square units}$
Step 2
Problem 4

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