The following integral can be evaluated only by reversing the order of integration. Sketch the region of integration, reverse the order of integration, and evaluate the integral. Integrate 0 to 27 Integrate cube root x to 3 (x/y^7+1) dy dx Choose the correct sketch of the region below. The reversed order of integration is Integrate Integrate (x/y^7 + 1) dx dy. (Simplify your answers.) The value of the integral is . (Type an exact answer.)
The Correct Answer and Explanation is :
To evaluate the integral
[
\int_{0}^{27} \int_{\sqrt[3]{x}}^{3} \left( \frac{x}{y^7} + 1 \right) dy \, dx,
]
we begin by sketching the region of integration in the xy-plane. The outer integral’s limits, ( x ) from 0 to 27, indicate that the region spans horizontally from ( x = 0 ) to ( x = 27 ). The inner integral’s limits, ( y ) from ( \sqrt[3]{x} ) to 3, suggest that for a fixed ( x ), ( y ) ranges from the curve ( y = \sqrt[3]{x} ) up to the horizontal line ( y = 3 ).
To visualize this, plot the curve ( y = \sqrt[3]{x} ), which passes through points like (0,0), (1,1), (8,2), and (27,3). The region of integration is bounded below by this curve, above by the line ( y = 3 ), on the left by ( x = 0 ), and on the right by ( x = 27 ).
Next, we reverse the order of integration. To do this, express the region in terms of ( y ) first and then ( x ). Observing the region, ( y ) ranges from 0 to 3. For a fixed ( y ), ( x ) ranges from ( x = 0 ) to ( x = y^3 ) (since ( x = y^3 ) is the inverse of ( y = \sqrt[3]{x} )). Thus, the reversed order of integration is:
[
\int_{0}^{3} \int_{0}^{y^3} \left( \frac{x}{y^7} + 1 \right) dx \, dy.
]
Now, evaluate the inner integral with respect to ( x ):
[
\int_{0}^{y^3} \left( \frac{x}{y^7} + 1 \right) dx.
]
This can be split into two integrals:
[
\int_{0}^{y^3} \frac{x}{y^7} \, dx + \int_{0}^{y^3} 1 \, dx.
]
Evaluate each separately:
- [
\int_{0}^{y^3} \frac{x}{y^7} \, dx = \frac{1}{y^7} \int_{0}^{y^3} x \, dx = \frac{1}{y^7} \left[ \frac{x^2}{2} \right]_{0}^{y^3} = \frac{1}{y^7} \cdot \frac{(y^3)^2}{2} = \frac{y^6}{2y^7} = \frac{1}{2y}.
] - [
\int_{0}^{y^3} 1 \, dx = \left[ x \right]_{0}^{y^3} = y^3.
]
Combining these results:
[
\int_{0}^{y^3} \left( \frac{x}{y^7} + 1 \right) dx = \frac{1}{2y} + y^3.
]
Now, integrate this result with respect to ( y ) from 0 to 3:
[
\int_{0}^{3} \left( \frac{1}{2y} + y^3 \right) dy.
]
Evaluate each term separately:
- [
\int_{0}^{3} \frac{1}{2y} \, dy = \frac{1}{2} \int_{0}^{3} \frac{1}{y} \, dy = \frac{1}{2} [\ln|y|]_{0}^{3}.
] As ( y ) approaches 0, ( \ln|y| ) approaches ( -\infty ), indicating an improper integral. However, since ( \frac{1}{y} ) is not integrable at ( y = 0 ), this term diverges.
Given the divergence in the ( \frac{1}{2y} ) term, the original integral does not converge to a finite value. Therefore, the value of the integral is divergent.
In summary, after reversing the order of integration and attempting to evaluate the integral, we find that the integral diverges due to the ( \frac{1}{2y} ) term, which is not integrable at ( y = 0 ).