Topics
>
Calculus and Integration>
Integration>
Areas Between Curves and the Axis>
Areas Between Curves and the Axis 2Areas Between Curves and the Axis 2
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 5 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- Answered
- Review
-
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
Help VideoCorrect
Correct!
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, 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
Help VideoCorrect
Great Work!
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, 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
Help VideoCorrect
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
Help VideoCorrect
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)`
Quizzes
- Indefinite Integrals 1
- Indefinite Integrals 2
- Indefinite Integrals 3
- Indefinite Integrals of Exponential Functions
- Indefinite Integrals of Logarithmic Functions 1
- Indefinite Integrals of Logarithmic Functions 2
- Indefinite Integrals of Trig Functions
- Definite Integrals
- Definite Integrals of Exponential Functions
- Definite Integrals of Logarithmic Functions
- Definite Integrals of Trig Functions
- Areas Between Curves and the Axis 1
- Areas Between Curves and the Axis 2
- Area Between Curves
- Volumes of Revolution 1
- Volumes of Revolution 2
- Volumes of Revolution 3
- Trapezoidal Rule
- Simpsons Rule
- Applications of Integration for Trig Functions