91Ó°ÊÓ

When solving double integrals, the order in which you integrate matters. Often, you'll find the need to integrate with respect to y first, depending on the given boundaries and the function itself. The process involves treating variables other than y, such as x, as constants. The function in our exercise, which is a polynomial in x, does not include y, so the integration with respect to y is straightforward. You multiply the function by the width of the strip, which is the difference between the upper and lower limits of y - in our case, from 0 to \(6x^2\). Imagine folding along the x-axis; each vertical strip is a constant 'height' determined by x.

For example, if you have a double integral where you're integrating \(x^3\) with respect to y from 0 to \(6x^2\), it's like asking how much volume is under the curve \(x^3\) stretched across a y-interval. As the value of y is not affecting the function, the result is simply the function multiplied by the y-interval, yielding \(x^3(6x^2)\). An essential tip for students: Always check if the function depends on the variable you're integrating with respect to - it often simplifies the calculation.

A fundamental technique in calculus is power rule integration, which serves as a cornerstone for solving polynomial integrations. The rule states that for any real number n, different from -1, the integral of \(x^n\) with respect to x is given by \(\frac{x^{n+1}}{n+1}\), plus a constant of integration in indefinite integrals. Applying this rule is like determining the 'antiderivative' of a function. When used in definite integrals, the constant of integration cancels out, and we evaluate the function at the upper and lower limits.

Let's apply this to our exercise where we need to integrate \(6x^5\). By the power rule, we increase the power by one to get \(6x^6\) and divide by the new power, yielding \(\frac{6x^6}{6}\). Simplifying gives us \(x^6\), which we will evaluate from 0 to 2. Remember that any number to the power of zero is one, so the lower limit will not contribute to our final answer. As an exercise improvement advice, make sure you carefully increase the power by one and equally divide by this new exponent to avoid common mistakes.

Once you've found the antiderivative of a function, evaluating definite integrals becomes a matter of substituting the upper and lower limits into this antiderivative. The process completes the calculation for the area under the curve or the accumulated quantity, depending on the context of the problem. In our problem, after applying the power rule of integration, we have the antiderivative \(x^6\). We then evaluate this at the upper limit of 2, and subtract the value at the lower limit of 0.

So, we have \(2^6\), which is 64, and since \(0^6\) is 0, our final step gives the result of 64. This value is the answer to our original double integral problem. It tells us the total volume under the 3-dimensional surface described by \(x^3\) over the given region. Finally, when evaluating an integral, remember this: always double-check your arithmetic when plugging in the limits, especially with higher powers, as even a small mistake could significantly change the outcome.