Integral Applications
Areas Between Curves and the Axis 1
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Question 1 of 5
1. Question
Find the area bounded by the curve `y=4−x^2` and lines `x=0` and `x=1`
- `\text(Area)=` (11/3) `\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
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$Identify the curve that the area is under and its bounds.`A` `=` $$\int (4-x^2) dx$$ The curve is `4-x^2` `A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (4-x^2) dx$$ The bounds are `x=0` and `x=1` Find the Indefinite Integral using the Power Rule$$\int (4-x^2) dx$$ `=` `4((x^(0+1))/(0+1)) – ((x^(2+1))/(2+1))` Apply the Power Rule `=` `4((x^1)/1) – (x^3)/3` Simplify `=` `4x – (x^3)/3` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (4-x^2) dx$$ `=` $$\left[4x – \frac{x^3}{3}\right]_{\color{#00880A}{0}}^{\color{#9a00c7}{1}}$$ `=` `[4(\color{#9a00c7}{1})-(\color{#9a00c7}{1}^3)/3] – [4(\color{#00880A}{0}) – (\color{#00880A}{0}^3)/3]` Substitute the upper `(1)` and lower limits `(0)` `=` `[4-1/3] – [0-0]` Simplify `=` `[12/3-1/3] – 0` `=` `11/3` `11/3 \text(square units)` -
Question 2 of 5
2. Question
Find the area bounded by the curve `y=2x^2-2` and lines `x=-1` and `x=1`
- `\text(Area)=` (8/3) `\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
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$Identify the curve that the area is under and its bounds.`A` `=` $$\int (2x^2-2) dx$$ The curve is `2x^2-2` `A` `=` $$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}} (2x^2-2) dx$$ The bounds are `x=-1` and `x=1` Find the Indefinite Integral using the Power Rule$$\int (2x^2-2) dx$$ `=` `2((x^(2+1))/(2+1)) – 2((x^(0+1))/(0+1))` Apply Power Rule `=` `2((x^3)/3) – 2((x^1)/1)` Simplify `=` `2/3 x^3 – 2x` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}} (2x^2-2) dx$$ `=` $$\left[\frac {2}{3} x^3 – 2x \right]_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$ `=` `[2/3(\color{#9a00c7}{1})^3-2(\color{#9a00c7}{1})] – [2/3(\color{#00880A}{-1})^3 – 2(\color{#00880A}{-1})]` Substitute the upper `(1)` and lower limits `(-1)` `=` `[2/3(1)-2] – [2/3(-1)+2]` Simplify `=` `[2/3-6/3] – [-2/3+6/3]` `=` `-4/3 – 4/3` `=` `-8/3` `=` `8/3` Get the absolute value `8/3 \text(square units)` -
Question 3 of 5
3. Question
Find the area bounded by the curve `y=x^2-7x+10` and lines `x=2` and `x=5`
- `\text(Area)=` (9/2) `\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
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$Identify the curve that the area is under and its bounds.`A` `=` $$\int (x^2-7x+10) dx$$ The curve is `x^2-7x+10` `A` `=` $$\int_{\color{#00880A}{2}}^{\color{#9a00c7}{5}} (x^2-7x+10) dx$$ The bounds are `x=2` and `x=5` Find the Indefinite Integral using the Power Rule$$\int (x^2-7x+10) dx$$ `=` `(x^(2+1)/(2+1)) -7 ((x^(1+1))/(1+1)) + 10((x^(0+1))/(0+1))` Apply Power Rule `=` `(x^3)/3 -7((x^2)/2) + 10((x^1)/1)` Simplify `=` `x^3/3 – (7x^2)/2 + 10x` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{2}}^{\color{#9a00c7}{5}} (x^2-7x+10) dx$$ `=` $$\left[\frac {x^3}{3} – \frac {7x^2}{2} + 10x \right]_{\color{#00880A}{2}}^{\color{#9a00c7}{5}}$$ `=` `[(\color{#9a00c7}{5}^3)/3 – 7((\color{#9a00c7}{5}^2)/2) +10(\color{#9a00c7}{5})] – [(\color{#00880A}{2}^3)/3 – 7((\color{#00880A}{2}^2)/2) + 10(\color{#00880A}{2})]` Substitute the upper `(5)` and lower limits `(2)` `=` `[125/3 -7((25)/2) +50] – [8/3-7((4)/2) + 20]` Simplify `=` `[125/3-175/2 + 50] – [8/3 -14 + 20]` `=` `25/6 – 26/3` `=` `-9/2` `=` `9/2` Get the absolute value `9/2 \text(square units)` -
Question 4 of 5
4. Question
Find the area bounded by the curve `y=1/2 x^3 + 1` and lines `x=2` and `x=0`
- `\text(Area)=` (4) `\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
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$Identify the curve that the area is under and its bounds.`A` `=` $$\int \left (\frac {1}{2} x^3 +1 \right) dx$$ The curve is `1/2 x^3+1` `A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} \left (\frac {1}{2} x^3 +1 \right) dx$$ The bounds are `x=0` and `x=2` Find the Indefinite Integral using the Power Rule$$\int \left (\frac {1}{2} x^3 +1 \right) dx$$ `=` `1/2((x^(3+1))/(3+1)) + 1((x^(0+1))/(0+1))` Apply Power Rule `=` `1/2((x^4)/4) + x` Simplify `=` `(x^4)/8 +x` Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} \left (\frac {1}{2} x^3 +1 \right) dx$$ `=` $$\left[\frac {x^4}{8} + x \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$$ `=` `[(\color{#9a00c7}{2}^4)/8+\color{#9a00c7}{2}] – [(\color{#00880A}{0}^4)/8 + \color{#00880A}{0}]` Substitute the upper `(2)` and lower limits `(0)` `=` `[16/8 +2] – [0]` Simplify `=` `2+2` `=` `4` `4 \text(square units)` -
Question 5 of 5
5. Question
Find the area bounded by the curve `y=x^3 ` and lines `x=-1` and `x=2`
- `\text(Area)=` (17/4) `\text(square units)`
Hint
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Keep Going!
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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
$$\int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} f(x) dx=\left[F(x)\right]_{\color{#00880A}{a}}^{\color{#9a00c7}{b}}=F(\color{#9a00c7}{b})-F(\color{#00880A}{a})$$Identify the curve that the area is under and its bounds.`A` `=` `A_1 + A_2` Divide the area into two `A` `=` $$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{0}} (x^3) dx + \int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} (x^3) dx$$ The bounds are `x=-1` and `x=2` Find the Indefinite Integral using the Power Rule$$\int (x^3) dx$$ `=` `(x^(3+1))/(3+1)` Apply Power Rule `=` `(x^4)/4` Simplify Find the Definite Integral`A` `=` $$\int_{\color{#00880A}{-1}}^{\color{#9a00c7}{0}} (x^3) dx + \int_{\color{#00880A}{0}}^{\color{#9a00c7}{2}} (x^3) dx$$ `=` $$\left[\frac {x^4}{4} \right]_{\color{#00880A}{-1}}^{\color{#9a00c7}{0}} + \left[\frac {x^4}{4} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{2}}$$ `=` `[((\color{#9a00c7}{0}^4)/4) – (((\color{#00880A}{-1})^4)/4)] + [((\color{#9a00c7}{2}^4)/4) – ((\color{#00880A}{0}^4)/4)] ` Substitute the upper and lower limits `=` `1/4 + 16/4` Simplify `=` `17/4` `17/4 \text(square units)`