Find the area between the x -axis and y = x 3 when 2 ≤ x ≤ 5.
44 1/3 square units
152 1/4 square units
39 square units
160 1/4 square units
area = ∫[2,5] x^3 dx = 1/4 x^4 [2,5] = ?
To find the area between the x-axis and the curve y = x^3 between 2 ≤ x ≤ 5, you can use definite integration.
The formula for finding the area between a curve and the x-axis is given by:
Area = ∫[a, b] |f(x)| dx
Where a and b are the limits of integration, and |f(x)| represents the absolute value of the function.
In this case, the function is y = x^3, and the limits of integration are 2 and 5. So the area can be calculated using the following integral:
Area = ∫[2, 5] |x^3| dx
To solve the integral, let's first consider the function x^3. As this is an odd function (raising x to the power of an odd number), the absolute value |x^3| is equivalent to x^3 when x is positive, and -x^3 when x is negative.
∫[2, 5] |x^3| dx = ∫[2, 5] x^3 dx
Now, we can integrate x^3 with respect to x:
= 1/4 * x^4 |[2, 5]
= 1/4 * (5^4 - 2^4)
= 1/4 * (625 - 16)
= 1/4 * 609
= 152.25
Thus, the area between the x-axis and y = x^3 when 2 ≤ x ≤ 5 is approximately 152 1/4 square units. Therefore, the correct option is 152 1/4 square units.