Area Between Two Curves

Question 1 of 4

Find the exact area bounded by the curves

$y=x^2$ and

$y=3x$

$\text (Area) =$ (9/2) $\text (square units)$

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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

Find the exact area bounded by the curves

$y=x-1$ and

$y=x^2-6x+5$

$\text (Area)=$ (125/6) $\text(square units)$

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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

Find the exact area bounded by the curves

$y=10-x^2$ and

$y=4-x$

$\text (Area)=$ (125/6) $\text(square units)$

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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

Find the exact area bounded by the curves

$y=x^2-6x+8$ and

$y=-x^2+4x-4$

$\text (Area)=$ (29/3) $\text(square units)$

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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

$A$	$=$	$29/3 \text(square units)$

$29/3 \text (square units)$

Area Between Two Curves

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