The Area Between Curves
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
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
- Answered
- Review
-
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
Help VideoCorrect
Great Work!
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=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)`
Quizzes
- Indefinite Integrals 1
- Indefinite Integrals 2
- Indefinite Integrals 3
- Definite Integrals
- Areas Between Curves and the Axis 1
- Areas Between Curves and the Axis 2
- The Area Between Curves
- Volumes of Revolution 1
- Volumes of Revolution 2
- Volumes of Revolution 3
- Trapezoidal Rule
- Simpsons Rule
- Integral of a Trigonometric Function 1
- Integral of a Trigonometric Function 2
- Applications of Integration for Trig Functions