Areas Between Curves and the Axis 2

Question 1 of 5

Find the area bounded by the curve

y = x (x - 1) (x - 3)

$y=x(x-1)(x-3)$ and lines

x = 0

$x=0$ and

x = 3

$x=3$

$Area =$ $\text(Area)=$ (37/12) $square units$ $\text(square units)$

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Integrate the function using the power rule to find

F (x)

$F(x)$ . Then to solve for the area between the curve and the axis, use the Definite Integral formula.

Definite Integral Formula

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$

First, simplify the polynomial.

$y$ $y$	$=$ $=$	$x (x - 1) (x - 3)$ $x(x-1)(x-3)$
	$=$ $=$	$x (x^{2} - 4 x + 3)$ $x(x^2-4x+3)$
$y$ $y$	$=$ $=$	$x^{3} - 4 x^{2} + 3 x$ $x^3-4x^2+3x$

Identify the curve that the area is under and its bounds.

$A$ $A$	$=$ $=$	$A_{1}$ $A_1$ $+$ $+$ $A_{2}$ $A_2$	Divide the area into two
$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (x^3-4x^2+3x) dx + \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}} (x^3-4x^2+3x) dx$	The bounds are $x = 0$ $x=0$ and $x = 3$ $x=3$

Find the Indefinite Integral using the Power Rule

$\int (x^3-4x^2+3x) dx$	$=$ $=$	$\frac{x^{3 + 1}}{3 + 1} - 4 (\frac{x^{2 + 1}}{2 + 1}) + 3 (\frac{x^{1 + 1}}{1 + 1})$ $(x^(3+1))/(3+1) – 4((x^(2+1))/(2+1)) + 3((x^(1+1))/(1+1))$	Apply Power Rule

	$=$ $=$	$\frac{x^{4}}{4} - 4 (\frac{x^{3}}{3}) + (3 \frac{x^{2}}{2})$ $(x^4)/4 – 4((x^3)/3) + (3(x^2)/2)$	Simplify

	$=$ $=$	$\frac{x^{4}}{4} - \frac{4}{3} x^{3} + \frac{3}{2} x^{2}$ $(x^4)/4 – 4/3x^3 + 3/2 x^2$

Calculate

A_{1}

$A_1$ using the Definite Integral

$A_{1}$ $A_1$	$=$ $=$	$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} \left(\frac{x^4}{4} – \frac{4}{3}x^3 + \frac{3}{2} x^2 \right) dx$

	$=$ $=$	$\left[\frac {x^4}{4} – \frac {4x^3}{3} + \frac {3x^2}{2} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{1}}$

	$=$ $=$	$[\frac{1^{4}}{4} - \frac{4 (1^{3})}{3} + \frac{3 (1^{2})}{2}] - [\frac{0^{4}}{4} - \frac{4 (0^{3})}{3} + \frac{3 (0^{2})}{2}]$ $[(\color{#9a00c7}{1}^4)/4 – (4(\color{#9a00c7}{1}^3))/3 + (3(\color{#9a00c7}{1}^2))/2] – [(\color{#00880A}{0}^4)/4 – (4(\color{#00880A}{0}^3))/3 + (3(\color{#00880A}{0}^2))/2]$	Substitute the upper and lower limits

	$=$ $=$	$[\frac{1}{4} - \frac{4}{3} + \frac{3}{2}] - [0 + 0 + 0]$ $[1/4 – 4/3 + 3/2] – [0+0+0]$	Simplify

$A_{1}$ $A_1$	$=$ $=$	$\frac{5}{12}$ $5/12$

Calculate

A_{2}

$A_2$ using the Definite Integral

$A_{2}$ $A_2$	$=$ $=$	$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}} \left(\frac{x^4}{4} – \frac{4}{3}x^3 + \frac{3}{2} x^2 \right) dx$

	$=$ $=$	$\left[\frac {x^4}{4} – \frac {4x^3}{3} + \frac {3x^2}{2} \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$

	$=$ $=$	$[\frac{3^{4}}{4} - \frac{4 (3^{3})}{3} + \frac{3 (3^{2})}{2}] - [\frac{1^{4}}{4} - \frac{4 (1^{3})}{3} + \frac{3 (1^{2})}{2}]$ $[(\color{#9a00c7}{3}^4)/4 – (4(\color{#9a00c7}{3}^3))/3 + (3(\color{#9a00c7}{3}^2))/2] – [(\color{#00880A}{1}^4)/4 – (4(\color{#00880A}{1}^3))/3 + (3(\color{#00880A}{1}^2))/2]$	Substitute the upper and lower limits

	$=$ $=$	$[\frac{81}{4} - \frac{4 (27)}{3} + \frac{3 (9)}{2}] - [\frac{1}{4} - \frac{4}{3} + \frac{3}{2}]$ $[81/4 – (4(27))/3 + (3(9))/2] – [1/4 – 4/3 + 3/2]$	Simplify

	$=$ $=$	$[\frac{81}{4} - \frac{108}{3} + \frac{27}{2}] - [\frac{5}{12}]$ $[81/4 – 108/3 + 27/2] – [5/12]$

	$=$ $=$	$- \frac{8}{3}$ $-8/3$

$A_{2}$ $A_2$	$=$ $=$	$\frac{8}{3}$ $8/3$	Use the absolute value for the area

Add the two areas.

$A$ $A$	$=$ $=$	$A_{1}$ $A_1$ $+$ $+$ $A_{2}$ $A_2$
	$=$ $=$	$\frac{5}{12}$ $5/12$ $+$ $+$ $\frac{8}{3}$ $8/3$	Substitute calculated values

$A$ $A$	$=$ $=$	$\frac{37}{12}$ $37/12$

\frac{37}{12} square units

$37/12 \text(square units)$

Question 2 of 5

Find the area bounded by the curve

x = (y - 2) (y - 1)

$x=(y-2)(y-1)$ and lines

y = 1

$y=1$ and

y = 2

$y=2$

$Area =$ $\text(Area)=$ (1/6) $square units$ $\text(square units)$

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Integrate the function using the power rule to find

F (x)

$F(x)$ . Then to solve for the area between the curve and the axis, use the Definite Integral formula.

Definite Integral Formula

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$

First, simplify the equation of the curve.

$x$ $x$	$=$ $=$	$(y - 2) (y - 1)$ $(y-2)(y-1)$

$x$ $x$	$=$ $=$	$y^{2} - 3 y + 2$ $y^2-3y+2$

Identify the bounds of the curve.

$A$ $A$	$=$ $=$	$\int (y^2-3y+2) dy$	The curve is $y^{2} - 3 y + 2$ $y^2-3y+2$

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{2}} (y^2-3y+2) dy$	The bounds are $y = 1$ $y=1$ and $y = 2$ $y=2$

Find the Indefinite Integral using the Power Rule

$\int (y^2-3y+2) dy$	$=$ $=$	$(\frac{y^{2 + 1}}{2 + 1}) - 3 (\frac{y^{1 + 1}}{1 + 1}) + 2 (\frac{y^{0 + 1}}{0 + 1})$ $((y^(2+1))/(2+1)) – 3((y^(1+1))/(1+1)) + 2((y^(0+1))/(0+1))$	Apply Power Rule

	$=$ $=$	$\frac{y^{3}}{3} - \frac{3 y^{2}}{2} + \frac{2 y}{1}$ $(y^3)/3 – (3y^2)/2 +(2y)/1$	Simplify

	$=$ $=$	$\frac{y^{3}}{3} - \frac{3 y^{2}}{2} + 2 y$ $(y^3)/3 – (3y^2)/2 +2y$

Find the Definite Integral

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{2}} (y^2-3y+2) dy$

	$=$ $=$	$\left[\frac {y^3}{3} – \frac {3y^2}{2} +2y \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{2}}$

	$=$ $=$	$[\frac{2^{3}}{3} - \frac{3 {(2)}^{2}}{2} + 2 (2)] - [\frac{1^{3}}{3} - \frac{3 {(1)}^{2}}{2} + 2 (1)]$ $[(\color{#9a00c7}{2}^3)/3 – (3(\color{#9a00c7}{2})^2)/2 +2(\color{#9a00c7}{2})] – [(\color{#00880A}{1}^3)/3 – (3(\color{#00880A}{1})^2)/2 +2(\color{#00880A}{1})]$	Substitute the upper $(2)$ $(2)$ and lower limits $(1)$ $(1)$

	$=$ $=$	$[\frac{8}{3} - 6 + 4] - [\frac{1}{3} - \frac{3}{2} + 2]$ $[8/3 – 6 +4] – [1/3 – 3/2 +2]$	Simplify

	$=$ $=$	$\frac{2}{3} - \frac{5}{6}$ $2/3-5/6$

	$=$ $=$	$- \frac{1}{6}$ $-1/6$

	$=$ $=$	$\frac{1}{6}$ $1/6$	Get the absolute value

\frac{1}{6} square units

$1/6 \text(square units)$

Question 3 of 5

Find the area bounded by the curve

y = \sqrt{x - 1}

$y=sqrt(x-1)$ and lines

y = 1

$y=1$ and

y = 5

$y=5$

$Area =$ $\text(Area)=$ (136/3) $square units$ $\text(square units)$

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Remember

Integrate the function using the power rule to find

F (x)

$F(x)$ . Then to solve for the area between the curve and the axis, use the Definite Integral formula.

Definite Integral Formula

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$

First, solve for

x

$x$ .

$y$ $y$	$=$ $=$	$\sqrt{x - 1}$ $sqrt(x-1)$

$y^{2}$ $y^2$	$=$ $=$	${(\sqrt{x - 1})}^{2}$ $(sqrt(x-1))^2$	Square both sides
$y^{2}$ $y^2$	$=$ $=$	$x - 1$ $x-1$
$y^{2}$ $y^2$ $+ 1$ $+1$	$=$ $=$	$x - 1$ $x-1$ $+ 1$ $+1$	Add $1$ $1$ to both sides
$y^{2} + 1$ $y^2+1$	$=$ $=$	$x$ $x$
$x$ $x$	$=$ $=$	$y^{2} + 1$ $y^2+1$

Identify the bounds of the curve.

$A$ $A$	$=$ $=$	$\int (y^2+1) dy$	The curve is $y^{2} + 1$ $y^2+1$

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{5}} (y^2+1) dy$	The bounds are $y = 1$ $y=1$ and $y = 5$ $y=5$

Find the Indefinite Integral using the Power Rule

$\int (y^2+1) dy$	$=$ $=$	$(\frac{y^{2 + 1}}{2 + 1}) + 1 (\frac{y^{0 + 1}}{0 + 1})$ $((y^(2+1))/(2+1)) + 1((y^(0+1))/(0+1))$	Apply Power Rule

	$=$ $=$	$\frac{y^{3}}{3} + y$ $(y^3)/3 +y$	Simplify

Find the Definite Integral

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{5}} (y^2+1) dy$

	$=$ $=$	$\left[\frac {y^3}{3} + y \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{5}}$

	$=$ $=$	$[\frac{5^{3}}{3} + 1 (5)] - [\frac{1^{3}}{3} + 1 (1)]$ $[(\color{#9a00c7}{5}^3)/3 +1(\color{#9a00c7}{5})] – [(\color{#00880A}{1}^3)/3 +1(\color{#00880A}{1})]$	Substitute the upper $(5)$ $(5)$ and lower limits $(1)$ $(1)$

	$=$ $=$	$[\frac{125}{3} + 5] - [\frac{1}{3} + 1]$ $[125/3 +5] – [1/3 +1]$	Simplify

	$=$ $=$	$\frac{140}{3} - \frac{4}{3}$ $140/3-4/3$

	$=$ $=$	$\frac{136}{3}$ $136/3$

\frac{136}{3} square units

$136/3 \text(square units)$

Question 4 of 5

Find the area bounded by the curve

x = 3 + 2 y - y^{2}

$x=3+2y-y^2$ and lines

y = - 1

$y=-1$ and

y = 3

$y=3$

$Area =$ $\text(Area)=$ (32/3) $square units$ $\text(square units)$

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Integrate the function using the power rule to find

F (x)

$F(x)$ . Then to solve for the area between the curve and the axis, use the Definite Integral formula.

Definite Integral Formula

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$

Identify the bounds of the curve.

$A$ $A$	$=$ $=$	$\int (3+2y-y^2) dy$	The curve is $3 + 2 y - y^{2}$ $3+2y-y^2$

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{3}} (3+2y-y^2) dy$	The bounds are $y = - 1$ $y=-1$ and $y = 3$ $y=3$

Find the Indefinite Integral using the Power Rule

$\int (3+2y-y^2) dy$	$=$ $=$	$(\frac{3 y^{0 + 1}}{0 + 1}) + 2 (\frac{y^{1 + 1}}{1 + 1}) - \frac{y^{2 + 1}}{2 + 1}$ $((3y^(0+1))/(0+1)) + 2((y^(1+1))/(1+1)) – (y^(2+1))/(2+1)$	Apply Power Rule

	$=$ $=$	$3 y + \frac{2 y^{2}}{2} - \frac{y^{3}}{3}$ $3y+(2y^2)/2 – (y^3)/3$	Simplify

	$=$ $=$	$3 y + y^{2} - \frac{y^{3}}{3}$ $3y+ y^2 – (y^3)/3$

Find the Definite Integral

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{3}} (3+2y-y^2) dy$

	$=$ $=$	$\left[3y+ y^2 – \frac {y^3}{3} \right]_{\color{#00880A}{-1}}^{\color{#9a00c7}{3}}$

	$=$ $=$	$[3 (3) + 3^{2} - \frac{3^{3}}{3}] - [3 (- 1) + {(- 1)}^{2} - \frac{{(- 1)}^{3}}{3}]$ $[3(\color{#9a00c7}{3}) + \color{#9a00c7}{3}^2 – (\color{#9a00c7}{3}^3)/3] – [3(\color{#00880A}{-1}) + (\color{#00880A}{-1})^2- ((\color{#00880A}{-1})^3)/3]$	Substitute the upper $(3)$ $(3)$ and lower limits $(- 1)$ $(-1)$

	$=$ $=$	$[9 + 9 - 9] - [- 3 + 1 + \frac{1}{3}]$ $[9+9-9] – [-3+1+1/3]$	Simplify

	$=$ $=$	$9 - (- \frac{5}{3})$ $9-(-5/3)$

	$=$ $=$	$\frac{32}{3}$ $32/3$

\frac{32}{3} square units

$32/3 \text(square units)$

Question 5 of 5

Find the area bounded by the curve

y = x^{3}

$y=x^3$ and lines

y = - 8

$y=-8$ and

y = - 1

$y=-1$

$Area =$ $\text(Area)=$ (45/4) $square units$ $\text(square units)$

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Integrate the function using the power rule to find

F (x)

$F(x)$ . Then to solve for the area between the curve and the axis, use the Definite Integral formula.

Definite Integral Formula

\int_{a}^{b} f (x) d x = {[F (x)]}_{a}^{b} = F (b) - F (a)

$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$

First, solve for

x

$x$ .

$y$ $y$	$=$ $=$	$x^{3}$ $x^3$

$y^{\frac{1}{3}}$ $y^(1/3)$	$=$ $=$	${(x^{3})}^{\frac{1}{3}}$ $(x^3)^(1/3)$	Raise both sides to a power of $\frac{1}{3}$ $1/3$
$y^{\frac{1}{3}}$ $y^(1/3)$	$=$ $=$	$x$ $x$
$x$ $x$	$=$ $=$	$y^{\frac{1}{3}}$ $y^(1/3)$

Identify the bounds of the curve.

$A$ $A$	$=$ $=$	$\int (y^{1/3}) dy$	The curve is $y^{\frac{1}{3}}$ $y^(1/3)$

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{-8}}^{\color{#9a00c7}{-1}} (y^{1/3}) dy$	The bounds are $y = - 8$ $y=-8$ and $y = - 1$ $y=-1$

Find the Indefinite Integral using the Power Rule

$\int (y^{1/3}) dy$	$=$ $=$	$(\frac{y^{\frac{1}{3} + 1}}{\frac{1}{3} + 1})$ $((y^(1/3+1))/(1/3+1))$	Apply Power Rule

	$=$ $=$	$\frac{y^{\frac{4}{3}}}{\frac{4}{3}}$ $(y^(4/3))/(4/3)$	Simplify

	$=$ $=$	$\frac{3}{4} (y^{\frac{4}{3}})$ $3/4 (y^(4/3))$

Find the Definite Integral

$A$ $A$	$=$ $=$	$\int_{\color{#00880A}{-8}}^{\color{#9a00c7}{-1}} (y^{1/3}) dy$

	$=$ $=$	$\left[ \frac {3}{4} y^{\frac {4}{3}}\right]_{\color{#00880A}{-8}}^{\color{#9a00c7}{-1}}$

	$=$ $=$	$[\frac{3}{4} {(- 1)}^{\frac{4}{3}}] - [\frac{3}{4} {(- 8)}^{\frac{4}{3}}]$ $[3/4(\color{#9a00c7}{-1})^(4/3)] – [3/4(\color{#00880A}{-8})^(4/3)]$	Substitute the upper $(- 1)$ $(-1)$ and lower limits $(- 8)$ $(-8)$

	$=$ $=$	$[\frac{3}{4} (1)] - [\frac{3}{4} (16)]$ $[3/4(1)] – [3/4(16)]$	Simplify

	$=$ $=$	$\frac{3}{4} - 12$ $3/4-12$

	$=$ $=$	$- \frac{45}{4}$ $-45/4$

	$=$ $=$	$\frac{45}{4}$ $45/4$	Get the absolute value

\frac{45}{4} square units

$45/4 \text(square units)$

Areas Between Curves and the Axis 2

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