Question:

Find the area enclosed between the curves $\displaystyle y = x^2 + 7$ and $\displaystyle y = 2x^2 + 3$.

Attempt:

$\displaystyle [ y = x^2 + 7 ] \times 2$

$\displaystyle [ y = 2x^2 + 3 ]$

$\displaystyle 2y = 2x^2 + 14$

$\displaystyle y = 2x^2 + 3$

$\displaystyle y = 11$

$\displaystyle 11 = 2x^2 + 3$

$\displaystyle 11 - 3 = 2x^2$

$\displaystyle 8 = 2x^2$

$\displaystyle \frac{8}{2} = x^2$

$\displaystyle 4 = x^2$

$\displaystyle x = \sqrt{4}$

$\displaystyle x = 2$

$\displaystyle y = 11 , x = 2$

$\displaystyle y = x^2 +7$

$\displaystyle = \frac{x^{2+1}}{2+1} +7x$

$\displaystyle = \frac{x^3}{3} + 7x$

$\displaystyle = (\frac{1}{3}x^3 + 7x)^2_0$

$\displaystyle = (\frac{1}{3}\times2^3 + 7\times2) - (\frac{1}{3}\times0^3 + 7\times0)$

$\displaystyle = 16\frac{2}{3} - 0$

$\displaystyle = 16\frac{2}{3}$

$\displaystyle y = 2x^2 + 3$

$\displaystyle = \frac{2x^{2+1}}{2+1} + 3x$

$\displaystyle = \frac{2x^3}{3} + 3x$

$\displaystyle = (\frac{2}{3}x^3 + 3x)^2_0$

$\displaystyle = (\frac{2}{3}\times2^3 + 3\times2) - (\frac{2}{3}\times0^3 + 3\times0)$

$\displaystyle = 11\frac{1}{3}$

Enclosed Area = $\displaystyle 16\frac{2}{3} - 11\frac{1}{3}$

$\displaystyle = 5\frac{1}{3}$

Answer in the text book is $\displaystyle 10\frac{2}{3}$