Areas Between Curves and the Axis 1

Question 1 of 5

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

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

Find the area bounded by the curve

$y=x^2-7x+10$ and lines

$x=2$ and

$x=5$

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

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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

Find the area bounded by the curve

$y=1/2 x^3 + 1$ and lines

$x=2$ and

$x=0$

$\text(Area)=$ (4) $\text(square units)$

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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 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

Find the area bounded by the curve

$y=x^3$ and lines

$x=-1$ and

$x=2$

$\text(Area)=$ (17/4) $\text(square units)$

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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.

$A$	$=$	$A_1 + A_2$	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

	$=$	$17/4$

$17/4 \text(square units)$

Areas Between Curves and the Axis 1

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