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Integrate the function f(x) = (1/6)*x^3 + 1/(3*x^2) with respect to x, between x = 1 and x = 3^(1/2), giving your answer in the form a + b*3^(1/2) where a and b are constants to be determined.

We can imagine our function as a curve. We have x values on the x axis and f(x) values on the y axis of our graph.

The question is essentially asking us to find the area between the curve, the x axis and the lines x=1 and x=31/2. It is a definite integral.

To integrate our function we simply use the rule that the integral of xis [1/(n+1)]*xn+1.

To make the function easier to work with we can rewrite f(x) as (1/6)*x3 + (1/3)*x-2.

Using the rule above we can integrate this function easily, with respect to x, to get (1/24)*x4 - (1/3)*x-1. Call this new function g(x).

To check we have obtained the correct integral, we can simply differentiate this new function, g(x). If the differential of g(x) is equal to f(x) then we have got the correct integral. It is always smart to do this check when working with integration.

The next step of our solution is to integrate f(x) between the given limits: x = 1 and x = 31/2. To do this we evaluate g(x) at x = 31/2, by simply subbing in the value for x, and then subtract from this answer the value of g(x) at x = 1.

We should get [3/8 - (1/9)*31/2] - [-7/24].

The final part of our solution is to rearrange our answer to get it in the form a + b*31/2.

The final answer to the question is: 2/3 - (1/9)*31/2.

12 months ago

Answered by Daniel, an A Level Economics tutor with MyTutor

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