Find a area between a curve and x-axis

Question

y= 2x^4 - 32 x-axis>>> -2,2

I know the answer I derived is wrong,(but close), can anyone check my work and let me know where I slipped?

2x^5

------- -32x

5

2(2)^5

-------- -32x

5

64 -32(2)

--- -

5

64 -64

--- -

5

128

----

5

Answer 1

We must integrate the function f(y) = 2x⁴ - 32 between x = 2 and x = -2

or,

₂

∫ ( 2x⁴ - 32 ) dx

^-2

Just integrating is the "indefinite" integral, while with a range it's a "definite" integral.

This becomes [ ²/₅ x⁵ - 32x + C ]² - [ ²/₅ x⁵ - 32x + C ]^-2

The coefficient of the x⁵ is ²/₅ because 2/5 x 5 = 10/5 = 2 as required. Integrating results in a constant C which will be included in the indefinite integral, but will cancel out in the definite integration.

Substituting the required x values gives (note the difference here, as you need to calculate for 2 different x values, and be careful of signs):

[ ²/₅ (2)⁵ - 32(2) + C ] - [ ²/₅ (-2)⁵ - 32(-2) + C ]

=> ( ⁶⁴/₅ - 64 + C) - ( -⁶⁴/₅ + 64 + C )

at this point, you can see you have a term missing!

=> ¹²⁸/₅ -128 = ¹²⁸/₅ - ⁶⁴⁰/₅

=> -⁵¹²/₅ = -102.4

The negative value of the definite integral indicates that the vertex of the curve (and hence the area enclosed) is below the x axis. It's at (0,-32).

The area between the given function and the x axis, for the given range, is 102.4 units²