this is number 2, see the diagram below. we want the area of the grey shaded region.

What is the area of the region bounded by the curvesy=x^4 +x^3 – 2x+ 3 andy=x^3 + 4x^2 – 2x+ 3

first we want to draw a diagram (below) then we want to find our limits of integration. we do that by finding where the curves intersect.

so y = x^4 + x^3 - 2x + 3 i'll call this f(x) and y = x^3 + 4x^2 - 2x + 3 i'll call this g(x)

=> x^4 + x^3 - 2x + 3 = x^3 + 4x^2 - 2x + 3

=> x^4 - 4x^2 = 0

=> x^2(x^2 - 4) = 0

=> x^2(x + 2)(x - 2) = 0

=> x = 0, x = -2 or x = 2 ...these are our points of intersection.

now we find whish graph is higher in both the regions [-2,0] and [0,2]. in both these regions, g(x) is greater than or equal to f(x), again, see diagram, if you could not produce the diagram, just plug in numbers in these regions into both f(x) and g(x), which ever one yields the greatest value is the higher function.

note that if differente functions were higher in the different intervals you would have to split the integral. example, if g(x) > f(x) on interval [-2,0] and f(x) > g(x) on [0,2], then the area would be

A = int{g(x) - f(x)}dx on [-2,0] + int{f(x) - g(x)}dx on [0,2]

since g(x) is always greater, we can integrate over the entire interval with one function, namely, g(x) - f(x)

so A = int{g(x) - f(x)}dx

=> A = int{x^3 + 4x^2 - 2x + 3 - (x^4 + x^3 - 2x + 3)}dx

=> A = int{-x^4 + 4x^2}dx

=> A = [(-1/5)x^5 + (4/3)x^3] evaluated between 2 and -2

=> A = -32/5 + 32/3 - 32/5 + 32/3

=> A = 128/15