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After sketching the graph of this, I thought it would be necessary to evaluate two separate integrals and add them together. When you see the graph, and find the intersection point of the two graphs, you get the line x=4. This shows how these two integrals need to be dealt with separately. That left me with evaluating as follows:

$$\int_0^4\int_0^x x^2\hspace{2 mm} dydx + \int_4^8\int_0^{16/x} x^2\hspace{2 mm} dydx$$

After sketching the graph of this, I thought it would be necessary to evaluate two separate integrals and add them together. When you see the graph, and find the intersection point of the two graphs, you get the line x=4. This shows how these two integrals need to be dealt with separately. That left me with evaluating as follows:

$$\int_0^4\int_0^x x^2\hspace{2 mm} dydx + \int_4^8\int_0^{16/x} x^2\hspace{2 mm} dydx$$

(This is actually 15.1 #16) After sketching the graph of this, I thought it would be necessary to evaluate two separate integrals and add them together. When you see the graph, and find the intersection point of the two graphs, you get the line x=4. This shows how these two integrals need to be dealt with separately. That left me with evaluating as follows:

$$\int_0^4\int_0^x x^2\hspace{2 mm} dydx + \int_4^8\int_0^{16/x} x^2\hspace{2 mm} dydx$$

This actually turns out to be pretty easy to integrate, it's the set up that is tricky! Hope this helps...