Question

Calculate the area between two curves:

$y=x^2$
$y=x+1$.

Answer 1

From the graph we get

Here $y=x^2$ & $y=x+1$ intercept at two point

So, $y=x^2=x+1$

$\therefore x^2=x+1$

$\therefore x^2-x-1=0$ So, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\therefore x=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(1\left)\right(-1)}}{2a}=\frac{1+\sqrt{5}}{2}$

$\therefore x_2\frac{1+\sqrt{5}}{2}$ & $x_1=\frac{1-\sqrt{5}}{2}$

So, $I={\int }_{x_1}^{x_2}\left[\right(x+1)-x^2]dx$

$I={\int }_{\frac{1-\sqrt{5}}{2}}^{\frac{1+\sqrt{5}}{2}}(-x^2+x+1)dx$

$I={[-\frac{x^3}{3}+\frac{x^2}{2}+x]}_{\frac{1-\sqrt{5}}{2}}^{\frac{1+\sqrt{5}}{2}}$

$I=\{-\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^3}{3}+\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^2}{2}+\left(\frac{1+\sqrt{5}}{2}\right)\}-\{-\frac{{\left(\frac{1-\sqrt{5}}{2}\right)}^3}{3}+\frac{{\left(\frac{1-\sqrt{5}}{2}\right)}^2}{2}+\left(\frac{1-\sqrt{5}}{2}\right)\}$

$I=\frac{{\left(\frac{1-\sqrt{5}}{2}\right)}^3-{\left(\frac{1+\sqrt{5}}{2}\right)}^3}{3}\times \frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^2-{\left(\frac{1-\sqrt{5}}{2}\right)}^2}{2}$

$\left(\frac{1+\sqrt{5}}{2}\right)-\left(\frac{1-\sqrt{5}}{2}\right)$

[Here $a^3-b^3=(a-b{)}^2(a^2-ab+b^2)=(a-b{)}^3+3ab(a-b)$ & $a^2-b^2=(a+b)(a-b)$]

$I=\frac{{\left(\frac{(1-\sqrt{5})-(1+\sqrt{5}}{2}\right)}^3+3\left(\frac{1-\sqrt{5}}{2}\right)\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{(1-\sqrt{5})-(1+\sqrt{5})}{2}\right)}{3}$

$+\frac{\left(\frac{1+\sqrt{5}+1-\sqrt{5}}{2}\right)\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right)}{2}$

$+\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right)$

$=\frac{(-\sqrt{5}{)}^3+\frac{3}{4}(1-5\left)\right(-\sqrt{5})}{3}+\left(1\right)\left(\sqrt{5}\right)$

$+\left(\sqrt{5}\right)$

$=\frac{-5\sqrt{5}+3\left(\sqrt{5}\right)}{3}+2\sqrt{5}$

$=\frac{-2\sqrt{5}}{3}+2\sqrt{5}$

$=\frac{4\sqrt{5}}{3}$

$I=\left|\frac{4\sqrt{5}}{3}\right|=\frac{4\sqrt{5}}{3}$