Area Between Two Curves (Leaving Cert Mathematics): Revision Notes

Area Between Two 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.

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

Therefore:

$$A = \frac{1}{3} - \frac{1}{4} = :success[\frac{1}{12}]$$

Alternative Method

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

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

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

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

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

3. Subtract:

$$A = 16 - \frac{32}{3} = :success[\frac{16}{3}]$$

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Alternative Method: Subtract Before Evaluating Limits

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

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