Area of Regions Between Curves

Area of Regions Between Curves

Area of a Region Between Two Curves (Integrate with Respect to x)

• Definition: Area between two curves when integrating with respect to x.

  • If f(x) > g(x) on the interval $[a, b]$, the area $A$ of the region bounded by the graphs of $f$ and $g$ on $[a, b]$ is given by:

  $A = \int_{a}^{b} (f(x) - g(x)) dx$

  • Case 1: If $g(x) = 0$, then:

  $A = \int_{a}^{b} f(x) dx$

  • Case 2: If $g(x) \neq 0$, then:

  $A = \int_{a}^{b} (f(x) - g(x)) dx$

Examples

• Rectangle:

  • To find the area of a rectangle, integrate $f(x) = h$ from $0$ to $l$.

  A = $A=\int_0^{l}\!f\left(x\right)\,dx=\int_0^{l}\!h\,dx$

• Triangular Plate:

  • To find the area of a triangular plate, integrate from $-1$ to $1$.

  $A=\int_{-1}^1\!f\left(x\right)\,dx$

  $=\int_{-1}^0\left(1+x\right)\!\,dx+\int_0^1\!\left(1-x\right)\,dx$

  $=\left\lbrack x+\frac{x^2}{2}\right\rbrack-1^0+\left\lbrack x-\frac{x^2}{2}\right\rbrack0^1=1$

• Area Between Two Curves:

  • Find the area of the region bounded by $f(x) = 5 - x^2$ and $g(x) = x^2 - 3$.

    1. Find Intersections:

    $f(x) = g(x) \Rightarrow 5 - x^2 = x^2 - 3 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = -2, 2$

    2. Calculate Area:

    $A = \int<em>{-2}^{2} (5 - x^2) - (x^2 - 3) dx = \int</em>{-2}^{2} (8 - 2x^2) dx = 2 \int_{0}^{2} (8 - 2x^2) dx$

    *Because the function is even.*
    

    $= 2 \Big[8x - \frac{2}{3}x^3\Big]_{0}^{2} = \frac{64}{3}$

• Area in the First Quadrant:

  • Find the area of the region in the first quadrant bounded by $f(x) = x^{\frac{2}{3}}$ and $g(x) = x - 4$.

    1. Find Intersection:

    $x^{\frac{2}{3}} = x - 4 \Rightarrow x^2 = (x - 4)^3 \Rightarrow x = 8$

    *The intersection point is $(8, 4)$.*
    

    2. Calculate Area:

    $R = \int<em>{0}^{4} f(x) dx + \int</em>{4}^{8} f(x) - g(x) dx = \int<em>{0}^{4} x^{\frac{2}{3}} dx + \int</em>{4}^{8} x^{\frac{2}{3}} - (x - 4) dx$

    $= \Big[\frac{3}{5}x^{\frac{5}{3}}\Big]<em>{0}^{4} + \Big[\frac{3}{5}x^{\frac{5}{3}} - \frac{1}{2}x^2 + 4x\Big]</em>{4}^{8} = \frac{56}{5}$

Area of a Region Between Two Curves (Integrate with Respect to y)

• Definition: Area between two curves when integrating with respect to y.

  • If $f(y) \geq g(y)$ on the interval $[c, d]$, the area $A$ of the region bounded by the graphs $x = f(y)$ and $x = g(y)$ on $[c, d]$ is given by:

  $A = \int_{c}^{d} (f(y) - g(y)) dy$

Example (Revisited)

• Area in the First Quadrant (Integrating with respect to y):

  • Find the area of the region in the first quadrant bounded by $f(x) = x^{\frac{2}{3}}$ and $g(x) = x - 4$ by integrating with respect to $y$ from $[c, d] = [0, 4]$.

    1. Rewrite Functions in terms of y:

    $f(x) = x^{\frac{2}{3}} \Rightarrow y = x^{\frac{2}{3}} \Rightarrow y^{\frac{3}{2}} = x \Rightarrow f(y) = y^{\frac{3}{2}}$

    $g(x) = x - 4 \Rightarrow y = x - 4 \Rightarrow y + 4 = x \Rightarrow g(y) = y + 4$

    2. Calculate Area:

    $R = \int<em>{0}^{4} g(y) - f(y) dy = \int</em>{0}^{4} (y + 4) - y^{\frac{3}{2}} dy$

    $= \Big[\frac{y^2}{2} + 4y - \frac{2}{5}y^{\frac{5}{2}}\Big]_{0}^{4} = \frac{56}{5}$