Area Between Two Curves
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Question 1 of 4
1. Question
Find the exact area bounded by the curves `y=x^2` and `y=3x`
- `\text (Area) =` (9/2) `\text (square units)`
Hint
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Remember
By integrating the difference of two functions, the area can be obtained.Definite Integral Formula
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$We are asked to find the area between the curves `y=3x` and `y=x^2``A` `=` `int (``3x``-``x^2``) dx` Subtract the lower curve from the higher curve Take the points of intersection as the limits`A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}} (3x-x^2) dx$$ Points intersect at `x=0` and `x=3` Find the Indefinite Integral using the Power Rule$$\int (3x-x^2) dx$$ `=` $$3 \left(\frac{x^{1+1}}{1+1} \right)- \left(\frac{x^{2+1}}{2+1}\right )$$ Apply the Power Rule `=` `3((x^2)/2) – (x^3)/3` Simplify `=` `(3x^2)/2 – (x^3)/3` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}} (3x-x^2) dx$$ `=` $$\left[\frac{3x^2}{2} – \frac{x^3}{3}\right]_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$ `=` `[(3(\color{#9a00c7}{3}^2))/2-(\color{#9a00c7}{3}^3)/3] – [(3(\color{#00880A}{0}^2))/2 – (\color{#00880A}{0}^3)/3]` Substitute the upper `(3)` and lower limits `(0)` `=` `[27/2-27/3] – [0-0]` Simplify `=` `27/2 – 9` `=` `27/2 – 18/2` `=` `9/2` `9/2 \text (square units)` -
Question 2 of 4
2. Question
Find the exact area bounded by the curves `y=x-1` and `y=x^2-6x+5`
- `\text (Area)=` (125/6) `\text(square units)`
Hint
Help VideoCorrect
Correct!
Incorrect
Remember
By integrating the difference of two functions, the area can be obtained.Definite Integral Formula
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$We are asked to find the area between the curves `y=x-1` and `y=x^2-6x+5``A` `=` `int (``x-1``-(``x^2-6x+5``)) dx` Subtract the lower curve from the higher curve `=` `int (-6+7x-x^2) dx` Take the points of intersection as the limits`A` `=` $$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{6}} (-6+7x-x^2) dx$$ Points intersect at `x=1` and `x=6` Find the Indefinite Integral using the Power Rule$$\int (-6+7x-x^2) dx$$ `=` $$-6 \left(\frac{x^{0+1}}{0+1} \right)+ 7\left(\frac{x^{1+1}}{1+1}\right ) – \left(\frac{x^{2+1}}{2+1}\right )$$ Apply the Power Rule `=` `-6(x^1)/1 +7(x^2)/2 -(x^3)/3` Simplify `=` `-6x +(7x^2)/2 -(x^3)/3` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{1}}^{\color{#9a00c7}{6}} (-6+7x-x^2) dx$$ `=` $$\left[-6x + \frac{7x^2}{2} – \frac{x^3}{3}\right]_{\color{#00880A}{1}}^{\color{#9a00c7}{6}}$$ `=` `[-6(\color{#9a00c7}{6})+ (7(\color{#9a00c7}{6}^2))/2-(\color{#9a00c7}{6}^3)/3] – [-6(\color{#00880A}{1}) +(7(\color{#00880A}{1}^2))/2 – (\color{#00880A}{1}^3)/3]` Substitute the upper `(6)` and lower limits `(1)` `=` `[-36+(7(36))/2-216/3] – [-6+7/2-1/3]` Simplify `=` `18-(-17/6)` `=` `125/6` `125/6 \text (square units)` -
Question 3 of 4
3. Question
Find the exact area bounded by the curves `y=10-x^2` and `y=4-x`
- `\text (Area)=` (125/6) `\text(square units)`
Hint
Help VideoCorrect
Fantastic!
Incorrect
Remember
By integrating the difference of two functions, the area can be obtained.Definite Integral Formula
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$We are asked to find the area between the curves `y=10-x^2` and `y=4-x``A` `=` `int (``10-x^2``-(``4-x``)) dx` Subtract the lower curve from the higher curve `=` `int (-x^2+x+6) dx` Take the points of intersection as the limits`A` `=` $$\int_{\color{#00880A}{-2}}^{\color{#9a00c7}{3}} (-x^2+x+6) dx$$ Points intersect at `x=-2` and `x=3` Find the Indefinite Integral using the Power Rule$$\int (-x^2+x+6) dx$$ `=` $$-\left(\frac{x^{2+1}}{2+1} \right)+ \left(\frac{x^{1+1}}{1+1}\right ) +6 \left(\frac{x^{0+1}}{0+1}\right )$$ Apply the Power Rule `=` `-(x^3)/3 +(x^2)/2 +6(x^1)/1` Simplify `=` `-(x^3)/3+(x^2)/2 +6x` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{-2}}^{\color{#9a00c7}{3}} (-x^2+x+6) dx$$ `=` $$\left[-\frac{x^3}{3} +\frac {x^2}{2} +6x \right]_{\color{#00880A}{-2}}^{\color{#9a00c7}{3}}$$ `=` `[-(\color{#9a00c7}{3}^3)/3 + (\color{#9a00c7}{3}^2)/2 +6(\color{#9a00c7}{3})] – [-(\color{#00880A}{-2})^3/3 + (\color{#00880A}{-2})^2/2 +6(\color{#00880A}{-2})]` Substitute the upper `(3)` and lower limits `(-2)` `=` `[-9+9/2+18] – [8/3+2-12]` Simplify `=` `27/2 – (-22/3)` `=` `125/6` `125/6 \text (square units)` -
Question 4 of 4
4. Question
Find the exact area bounded by the curves `y=x^2-6x+8` and `y=-x^2+4x-4`
- `\text (Area)=` (29/3) `\text(square units)`
Hint
Help VideoCorrect
Excellent!
Incorrect
Remember
By integrating the difference of two functions, the area can be obtained.Definite Integral Formula
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$We are asked to find the area between the curves `y=x^2-6x+8` and `y=-x^2+4x-4``A` `=` `int (``x^2-6x+8``-(``-x^2+4x-4``)) dx` Subtract the lower curve from the higher curve `=` `int (2x^2-10x+12) dx` Take the points of intersection as the limits. We can divide the area into two.`A` `=` `A_1``+``A_2` `=` `int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} (2x^2-10x+12) dx`` +`` int_{\color{#00880A}{2}}^{\color{#9a00c7}{3}} (2x^2-10x+12) dx` Get individual integrals Find the Indefinite Integral using the Power Rule$$\int (2x^2-10x+12) dx$$ `=` $$\left(2 \frac{x^{2+1}}{2+1} \right)-10 \left(\frac{x^{1+1}}{1+1}\right ) +12 \left(\frac{x^{0+1}}{0+1}\right )$$ Apply the Power Rule `=` `(2x^3)/3 -(10x^2)/2 +(12(x^1))/1` Simplify `=` `(2x^3)/3-5x^2 +12x` Solve for `A_1` using Definite Integral`A_1` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} (2x^2-10x+12) dx$$ `=` $$\left[2\frac{x^3}{3}- 5x^2 +12x \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$$ `=` `[(2(\color{#9a00c7}{2}^3))/3 – 5(\color{#9a00c7}{2}^2) +12(\color{#9a00c7}{2})] – [(2(\color{#00880A}{0}^3))/3 – 5(\color{#00880A}{0}^2) +12(\color{#00880A}{0})]` Substitute the upper `(2)` and lower limits `(0)` `=` `[16/3 -20 +24] – [0]` Simplify `A_1` `=` `28/3 \text(square units)` Solve for `A_2`.`A_2` `=` $$\int_{\color{#00880A}{2}}^{\color{#9a00c7}{3}} (2x^2-10x+12) dx$$ `=` $$\left[2\frac{x^3}{3}- 5x^2 +12x \right]_{\color{#00880A}{2}}^{\color{#9a00c7}{3}}$$ `=` `[(2(\color{#9a00c7}{3}^3))/3 – 5(\color{#9a00c7}{3}^2) +12(\color{#9a00c7}{3})] – [(2(\color{#00880A}{2}^3))/3 – 5(\color{#00880A}{2}^2) +12(\color{#00880A}{2})]` Substitute the upper `(3)` and lower limits `(2)` `=` `[54/3 -45 +36] – [16/3 – 20 + 24]` Simplify `=` `9 – 28/3` `=` `-1/3` Get the absolute value `A_2` `=` `1/3 \text(square units)` Add the two areas`A` `=` `A_1``+``A_2` `=` `28/3``+``1/3` Substitute calculated values `A` `=` `29/3 \text(square units)` `29/3 \text (square units)`