Area between 2 curves (Edexcel A-Level Mathematics): Revision Notes

8.2.10 Area between 2 curves

Integration: Further Area

Area Between Two Curves

Find the area of the shaded region $A$.

Area between the uppermost curve and the $x$-axis, take the area between the lowermost curve and the $x$-axis.

$$A_1 = \int_0^1 x^2 \, dx = \left[ \frac{1}{3}x^3 \right]_0^1 = \left( \frac{1}{3}(1)^3 - \frac{1}{3}(0)^3 \right) = \frac{1}{3}$$ $$A_2 = \int_0^1 x^3 \, dx = \left[ \frac{1}{4}x^4 \right]_0^1 = \left( \frac{1}{4}(1)^4 - \frac{1}{4}(0)^4 \right) = \frac{1}{4}$$

Therefore:

$$A = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$$

Alternative Method

$$\int_0^1 x^2 - x^3 \, dx = \left[ \frac{1}{3}x^3 - \frac{1}{4}x^4 \right]_0^1 = \left( \frac{1}{3}(1)^3 - \frac{1}{4}(1)^4 \right) - \left( \frac{1}{3}(0)^3 - \frac{1}{4}(0)^4 \right) = \frac{1}{12}$$

---

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:

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

---

---

1. Find the area between the "upper curve" (red) and the x-axis:

$$\int_1^3 \left( 11 - \frac{9}{x^2} \right) dx = \int_1^3 11 - 9x^{-2} dx$$ $$= \left[ 11x + 9{x^{-1}} \right]_1^3$$ $$= (11(3) + 9(3)^{-1}) - (11(1) + 9(1)^{-1}) = 16$$

2. Find the area between the "lower curve" (blue) and the $x$-axis:

$$\int_1^3 \left( x^2 + 1 \right) dx = \left[ \frac{x^3}{3} + x \right]_1^3 = \left( \frac{3^3}{3} + 3 \right) - \left( \frac{1^3}{3} + 1 \right) = \frac{32}{3}$$

3. Subtract:

$$A = 16 - \frac{32}{3} = \frac{16}{3}$$

---

Alternative Method: Subtract Before Evaluating Limits

$$\int_1^3 11 - 9{x^{-2}} - (x^2 + 1)\ dx = \int_1^3 10 - 9{x^{-2}} - x^2 \ dx$$ $$= \left[ 10x + 9{x^{-1}} - \frac{x^3}{3} \right]_1^3$$ $$= \left( 10(3) + 9(3)^{-1} - \frac{3^3}{3} \right) - \left( 10(1) + 9(1)^{-1} - \frac{1^3}{3} \right) = \frac{16}{3}$$

---

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.

4. Region (A)

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

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

$$\text{Let } y = 0 \Rightarrow 0 = \frac{8}{x^2} - 2 \Rightarrow 0 = 8 - 2x^2 \Rightarrow 8 = 2x^2 =8 \Rightarrow x^2 = 4$$ $$\Rightarrow x = 2 \text{ or } x = -2 \, (\text{not valid since } x > 0)$$ $$\therefore \int_1^2 8{x^{-2}} - 2 \ dx = \left[- 8x^{-1}- 2x \right]_1^2 = \left( -8(2)^{-2} - 2(2) \right) - \left( -8(1)^{-1} - 2(1) \right)$$ $$= 2$$

6. Total Area:

$$\text{Area} = (A) + (B) = \frac{12}{5} + 2 = \frac{22}{5}$$

---

Integration: Area Between a Curve and the $y$-Axis

To find the area between a curve and the $x$-axis, we calculate:

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

To find the area between a curve and the $y$-axis, we calculate:

$$\int_{y_1}^{y_2} x \, dy \quad \text{(using y-limits, not x-limits)}$$

---

---

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 $x = 2 \Rightarrow y =3+\sqrt {2+2} = 5$

When $x = 14 \Rightarrow y = 3 + \sqrt {14+2} = 7$

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:

$$= \left(\frac{7^3}{3} - 3(7^2) + 7(7)\right) - \left(\frac{5^3}{3} - 3(5^2) + 7(5)\right)$$ $$= \frac{44}{3}$$

---