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 = 7x^2 , y = x^2 +1

7x^2 = x^2 +1
6x² = 1
x = ±√(1/6)

∫(x²+1-7x²) dx from -√(1/6) to √(1/6)
a)= 2 ∫(x²+1-7x²) dx from 0 to √(1/6)
= 2 ∫(1-6x²) dx from 0 to √(1/6)
= 2[x-2x³] from 0 to √(1/6)
= 2(1/√6 - 2/(√6)³)
= 2(1/√6 - 2/(6√6))
= 2/√6( 1 - 1/3)
= 4/(3√6)

Where did the 2 from the very first a came from?

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 = 7x^2 , y = x^2 +1

7x^2 = x^2 +1
6x² = 1
x = ±√(1/6)

∫(x²+1-7x²) dx from -√(1/6) to √(1/6)
a)= 2 ∫(x²+1-7x²) dx from 0 to √(1/6)
= 2 ∫(1-6x²) dx from 0 to √(1/6)
= 2[x-2x³] from 0 to √(1/6)
= 2(1/√6 - 2/(√6)³)
= 2(1/√6 - 2/(6√6))
= 2/√6( 1 - 1/3)
= 4/(3√6)

Where did the 2 from the very first a came from?

anonymous · Answer

the 2 came from the fact that the graph over that region is symmetrical
example
the integral of y=x^2 over the interval (-1,1) is symmetrical
so, instead of taking the area from -1 to 1, we take the area from 0 to 1 and multiply it by 2
this is usually easier because 0 is a limit of integration, and its less messy to put 0 into an equation

anonymous · Answer

Ah! Thanks so much etemplin!