Find the area between each curve and the x-axis for the given interval.

a. y=x^4+5 from x=0 to x=5
650

b. y=3x^2+5x-1 from x=1 to x=4
97.5

c. y=4x-x^3 from x=0 to x=2
4

all 3 are correct.

To find the area between a curve and the x-axis for a given interval, you can use integration. The definite integral of a function represents the area between the curve and the x-axis within the specified interval.

a. To find the area between the curve y = x^4 + 5 and the x-axis from x = 0 to x = 5, you can integrate the function with respect to x over the given interval:

∫(0 to 5) (x^4 + 5) dx

Integrating the function, we get:

[(1/5)x^5 + 5x] evaluated from 0 to 5

Substituting the upper and lower limits:

[(1/5)(5^5) + 5(5)] - [(1/5)(0^5) + 5(0)]
= (1/5)(3125) + 25 - 0
= 625 + 25
= 650

Therefore, the area between the curve y = x^4 + 5 and the x-axis from x = 0 to x = 5 is 650 square units.

b. To find the area between the curve y = 3x^2 + 5x - 1 and the x-axis from x = 1 to x = 4, we can integrate the function:

∫(1 to 4) (3x^2 + 5x - 1) dx

Integrating the function, we get:

[x^3 + (5/2)x^2 - x] evaluated from 1 to 4

Substituting the upper and lower limits:

[4^3 + (5/2)(4^2) - 4] - [1^3 + (5/2)(1^2) - 1]
= 64 + 40 - 4 - 1 - (5/2) + 1
= 99 - (5/2)
= 193/2
= 96.5

Therefore, the area between the curve y = 3x^2 + 5x - 1 and the x-axis from x = 1 to x = 4 is 97.5 square units.

c. To find the area between the curve y = 4x - x^3 and the x-axis from x = 0 to x = 2, we can integrate the function:

∫(0 to 2) (4x - x^3) dx

Integrating the function, we get:

[2x^2 - (1/4)x^4] evaluated from 0 to 2

Substituting the upper and lower limits:

[2(2^2) - (1/4)(2^4)] - [2(0^2) - (1/4)(0^4)]
= 8 - 4 - 0 + 0
= 4

Therefore, the area between the curve y = 4x - x^3 and the x-axis from x = 0 to x = 2 is 4 square units.