Question

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region y=4x^2, y=x^2 + 3

Answer 1

Answer:

Step-by-step explanation:

integrate with respect to x

Area is symmetric so will be twice the integral from zero to 1

( I cannot seem to figure out how to make the lower limit negative in this editor)

$A = 2\int\limits^1_0 {(x^2 + 3) - 4x^2} \, dx$)

$A = 2\int\limits^1_0 {-3x^2 + 3} \, dx$

$A = 6\int\limits^1_0 {-x^2 + 1} \, dx$

A = 6(-⅓x³ + x)$\left \| {{x=1} \atop {x=0}} \right.$

A = 6((-⅓(1)³ + (1)) - (-⅓(0)³ + (0)))

A = 6(⅔)

A = 4 units²