Area Between Curve and y-axis

Area Between a Curve and the $y$ -Axis

The formula for the area under a curve (i.e., between the curve and the $x$ -axis) is:

\int_{x_1}^{x_2} y \, dx

The formula for the area between a curve and the $y$ -axis is:

\int_{y_1}^{y_2} x \, dy

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Example: Find the shaded area in the following diagram:

Rearrange the equation to say $x = f(y)$

y = x^2 \Rightarrow x = \pm \sqrt{y} \Rightarrow x = \sqrt{y} \\ \text {since it's in the :highlight[positive quadrant].}

Find the $y$ -limits

x = 2 \Rightarrow y = 2^2 = 4

y = 4^2 = 16

Perform the integration

\int_{y_1}^{y_2} x \, dy = \int_4^{16} y^{\frac{1}{2}} \, dy = \left[ \frac{2}{3} y^{\frac{3}{2}} \right]_4^{16}

= \left( \frac{2}{3} (16)^{\frac{3}{2}} \right) - \left( \frac{2}{3} (4)^{\frac{3}{2}} \right) = \frac{128}{3} - \frac{16}{3} = :highlight[\frac{112}{3}]

\int_4^{16} \frac{1}{2} x^2 \, dx = \frac{112}{3}

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Q2.

(a) Find $\int (x^2 + 4)(x-6) \, dx$ .

(b)

The diagram shows the curve $y = 6x^{\frac{3}{2}}$ and part of the curve $y = \frac{8}{x^2} - 2$ , which intersect at the point $(1, 6)$ . Use integration to find the area of the shaded region enclosed by the two curves and the x-axis.

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Be careful in situations in which the area is not contained from above and below by two curves.

Region (A)

\int_0^1 6x^{\frac{3}{2}} \, dx = \left[ \frac{2}{5} \times 6x^{\frac{5}{2}} \right]_0^1 = \left[ \frac{12}{5} x^{\frac{5}{2}} \right]_0^1 = \left( \frac{12}{5}(1)\right) - \left( \frac{12}{5}(0) \right)

= :highlight[\frac{12}{5}]

Region (B) Root of $y = \frac{8}{x^2} - 2$ :

\text{Let } y = 0 \Rightarrow 0 = \frac{8}{x^2} - 2 \Rightarrow 0 = 8 - 2x^2 \Rightarrow 8 = 2x^2 =8 \Rightarrow x^2 = 4

\Rightarrow x = 2 \text{ or } x = -2 \, (\text{not valid since } x > 0)

\therefore \int_1^2 8{x^{-2}} - 2 \ dx = \left[- 8x^{-1}- 2x \right]_1^2 = \left( -8(2)^{-2} - 2(2) \right) - \left( -8(1)^{-1} - 2(1) \right)

= :highlight[2]

Total Area:

\text{Area} = (A) + (B) = \frac{12}{5} + 2 = :highlight[\frac{22}{5}]

infoNote

Example:

Find the shaded area for $y = x^3$ .

Rearrange the equation to the form $x = f(y)$ :

y = x^3 \Rightarrow x = y^{\frac{1}{3}}

Find the y-limits:

When $x = 2 \Rightarrow y = 2^3 = 8$
When $x = 3 \Rightarrow y = 3^3 = 27$

Calculate $\int_{y_1}^{y_2} x \, dy$ :

\int_8^{27} y^{\frac{1}{3}} \, dy = \left[ \frac{3}{4} y^{\frac{4}{3}} \right]_8^{27} = \frac{3}{4} \left( 27^{\frac{4}{3}} \right) - \frac{3}{4} \left( 8^{\frac{4}{3}} \right)= :highlight[\frac{195}{4}]

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Q1.

The diagram shows the curve $y = 3 + \sqrt{x+2}$ .

The shaded region is bounded by the curve, the $y$ -axis, and two lines parallel to the $x$ -axis which meet the curve where $x = 2$ and $x = 14$ .

Question :

(i) Show that the area of the shaded region is given by

\int_5^7 (y^2 - 6y + 7) \, dy.

(ii) Hence find the exact area of the shaded region.

Solution

(i) Start with the equation of the curve:

y = 3 + \sqrt{x+2}

Rearrange to solve for $x$ :

y - 3 = \sqrt{x+2} \Rightarrow (y-3)^2 = x+2 \Rightarrow x = (y-3)^2 - 2

=y^2-6y+9-2 = y^2-6y+7

When $x = 2 \Rightarrow y =3+\sqrt {2+2} = :highlight[5]$

When $x = 14 \Rightarrow y = 3 + \sqrt {14+2} = :highlight[7]$

Therefore Area

\int_5^7 y^2 - 6y + 7 \, dy

(ii) Calculate the area using the integral:

\int_5^7 (y^2 - 6y + 7) \, dy = \left[\frac{y^3}{3} - 3y^2 + 7y\right]_5^7

Substitute the limits $5$ and $7$ :

= \left(\frac{7^3}{3} - 3(7^2) + 7(7)\right) - \left(\frac{5^3}{3} - 3(5^2) + 7(5)\right)

= :highlight[\frac{44}{3}]

Area Between Curve and y-axis Simplified Revision Notes for Leaving Cert Mathematics

Area Between Curve and y-axis

Area Between a Curve and the $y$ -Axis

Example: Find the shaded area in the following diagram:

Q2.

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500K+ Students Use These Powerful Tools to Master Area Between Curve and y-axis For their Leaving Cert Exams.

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Area Between a Curve and the yyy-Axis

Example: Find the shaded area in the following diagram:

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500K+ Students Use These Powerful Tools to Master Area Between Curve and y-axis For their Leaving Cert Exams.

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