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.

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

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

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

Find the Indefinite Integral using the Power Rule

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

	$=$	$(x^4)/4 – 4((x^3)/3) + (3(x^2)/2)$	Simplify

	$=$	$(x^4)/4 – 4/3x^3 + 3/2 x^2$

Calculate

$A_1$ using the Definite Integral

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

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

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

	$=$	$[1/4 – 4/3 + 3/2] – [0+0+0]$	Simplify

$A_1$	$=$	$5/12$

Calculate

$A_2$ using the Definite Integral

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

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

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

	$=$	$[81/4 – (4(27))/3 + (3(9))/2] – [1/4 – 4/3 + 3/2]$	Simplify

	$=$	$[81/4 – 108/3 + 27/2] – [5/12]$

	$=$	$-8/3$

$A_2$	$=$	$8/3$	Use the absolute value for the area

Add the two areas.

$A$	$=$	$A_1$ $+$ $A_2$
	$=$	$5/12$ $+$ $8/3$	Substitute calculated values

$A$	$=$	$37/12$

$37/12 \text(square units)$

Question 2 of 5

Find the area bounded by the curve

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

$y=1$ and

$y=2$

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

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Integrate the function using the power rule to find

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

Definite Integral Formula

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

First, simplify the equation of the curve.

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

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

Identify the bounds of the curve.

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

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

Find the Indefinite Integral using the Power Rule

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

	$=$	$(y^3)/3 – (3y^2)/2 +(2y)/1$	Simplify

	$=$	$(y^3)/3 – (3y^2)/2 +2y$

Find the Definite Integral

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

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

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

	$=$	$[8/3 – 6 +4] – [1/3 – 3/2 +2]$	Simplify

	$=$	$2/3-5/6$

	$=$	$-1/6$

	$=$	$1/6$	Get the absolute value

$1/6 \text(square units)$

Question 3 of 5

Find the area bounded by the curve

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

$y=1$ and

$y=5$

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

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Remember

Integrate the function using the power rule to find

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

Definite Integral Formula

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

First, solve for

$x$ .

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

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

Identify the bounds of the curve.

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

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

Find the Indefinite Integral using the Power Rule

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

	$=$	$(y^3)/3 +y$	Simplify

Find the Definite Integral

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

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

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

	$=$	$[125/3 +5] – [1/3 +1]$	Simplify

	$=$	$140/3-4/3$

	$=$	$136/3$

$136/3 \text(square units)$

Question 4 of 5

Find the area bounded by the curve

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

$y=-1$ and

$y=3$

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

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Remember

Integrate the function using the power rule to find

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

Definite Integral Formula

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

Identify the bounds of the curve.

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

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

Find the Indefinite Integral using the Power Rule

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

	$=$	$3y+(2y^2)/2 – (y^3)/3$	Simplify

	$=$	$3y+ y^2 – (y^3)/3$

Find the Definite Integral

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

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

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

	$=$	$[9+9-9] – [-3+1+1/3]$	Simplify

	$=$	$9-(-5/3)$

	$=$	$32/3$

$32/3 \text(square units)$

Question 5 of 5

Find the area bounded by the curve

$y=x^3$ and lines

$y=-8$ and

$y=-1$

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

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Integrate the function using the power rule to find

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

Definite Integral Formula

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

First, solve for

$x$ .

$y$	$=$	$x^3$

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

Identify the bounds of the curve.

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

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

Find the Indefinite Integral using the Power Rule

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

	$=$	$(y^(4/3))/(4/3)$	Simplify

	$=$	$3/4 (y^(4/3))$

Find the Definite Integral

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

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

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

	$=$	$[3/4(1)] – [3/4(16)]$	Simplify

	$=$	$3/4-12$

	$=$	$-45/4$

	$=$	$45/4$	Get the absolute value

$45/4 \text(square units)$

Areas Between Curves and the Axis 2

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