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Areas Between Curves and the Axis>
Areas Between Curves and the Axis 2Areas Between Curves and the Axis 2
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Question 1 of 5
1. Question
Find the area bounded by the curve `y=x(x-1)(x-3) ` and lines `x=0` and `x=3`
- `\text(Area)=` (37/12) `\text(square units)`
Hint
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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 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
2. Question
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)`
Hint
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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, 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
3. Question
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)`
Hint
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Excellent!
Incorrect
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
4. Question
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)`
Hint
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Keep Going!
Incorrect
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
5. Question
Find the area bounded by the curve `y=x^3` and lines `y=-8` and `y=-1`
- `\text(Area)=` (45/4) `\text(square units)`
Hint
Help VideoCorrect
Well Done!
Incorrect
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` `=` `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)`