Using integration , find the area of the region enclosed by $ x^2=y $ , the line $ y=x+2 $ and the x-axis.

Question

My attempt :-

$$ \int_{-2}^{-1}{{x} \ +\ 2 }\,\ dx \ + \int_{-1}^{0}{{x} ^2}\, dx \\ $$

$$ {\frac{x^2}{2} \ +\ 2x\ } \biggr|_{-2}^{-1} \,\ dx \ + {\frac{x^3}{3 }\ } \biggr|_{-1}^{0} \, dx \\ $$

Which gives the answer as $$ \frac{5}{6} $$

But the answer as per book is $$\frac{9}{2} $$

What's the mistake in my solution ?

Answer 1

Your working is right, you found the area of the region enclosed by $ x^2=y $ , the line $ y=x+2 $ and the x-axis.

Unfortunately, the book intended to ask the region enclosed by $x^2=y$ and the line $y=x+2$.

We can see from the graph that the area is bounded above by $1$.

Answer 2

What you should compute is$$\int_{-1}^2x+2-x^2\,\mathrm dx.$$The values $x=-1$ and $x=2$ are those $x$'s such that $x+2=x^2$. Besides, not that$$x\in[-1,2]\implies x+2\geqslant x^2.$$