Answer to Question #223429 in Calculus for FRANCIS

Question

1. Find R R
D
(x + 2y)dA, where D is the region bounded by the parabolas
y = x
2
, and y = 1 + x

Answer 1

$exact \space question\\ evaluate \space \int \space \int_{D} \space (x+2y)dA\\ where \space D \space is \space the \space region \space bounded \space by \space the \space parabolas \space y=2x^2 \space and \space y=1+x^2\\ ------------------------\\ solution\\ diagram\\:- where \space blue \space line \space represent \space y=2x^2\\ \space red \space line \space represent \space y=1+x^2$

$\int \space \int_{D} \space (x+2y)dA=\int_{-1}^1 \space \int_{2x^2}^{1+x^2}(x+2y)dydx\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\int_{-1}^1 \space (xy+y^2)|_{2x^2}^{1+x^2}dx\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\int_{-1}^1[(x(1+x^2)+(1+x^2)^2)-(x(2x^2)+(2x^2)^2)]dx\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\int_{-1}^1[(x+x^3+1+2x^2+x^4)-(2x^3+4x^4)]dx\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\int_{-1}^1[(1+x+2x^2-x^3-3x^4)]dx\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =[x+\frac{x^2}{2}+\frac{2x^3}{3}-\frac{x^4}{4}-\frac{3x^5}{5}]_{-1}^1\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\frac{32}{15}\\ hence \space required \space solution=\frac{32}{15}\\$