Evaluating Definite Integrals: A Step-by-Step Guide

In the realm of integral calculus, evaluating definite integrals is a fundamental skill. In this article, we will walk through a detailed example, employing the substitution method to solve one such problem. Let's explore how to evaluate the definite integral:

Let I (int_0^8 frac{1}{1sqrt[3]{x}}dx)

Step-by-Step Solution

First, we introduce a substitution to simplify the integral:

Let (sqrt[3]{x} u)

Then (x u^3)

Differentiate both sides to find (dx)

(frac{dx}{du} 3u^2) hence, (dx 3u^2du)

Substitute (sqrt[3]{x}) and (dx) into the original integral:

I (int_0^2 frac{3u^2}{1cdot u}du int_0^2 3u du - int_0^2 frac{3u^{1-1}cdot du}{1cdot u})

Separate the integral into two parts:

First Part

I (int_0^2 3u du) ([frac{3u^2}{2}]_0^2 frac{3cdot2^2}{2} - frac{3cdot0^2}{2} 6 - 0 6)

Second Part

I (- int_0^2 frac{3du}{1u} - [3cdot ln |u|]_0^2 -[3cdot ln (2)] - [3cdot ln (0)])

Note that the integral from 0 to 2 of (frac{1}{u} du) evaluates to (ln left|uright|) and the term involving (ln (0)) is typically undefined, but for our bounds, the integral evaluates nicely.

Final Integration

Combining the results from both parts, we obtain:

I 6 - 3ln(2) 3ln(3))

Alternative Methods and Verification

Another method to verify this solution is to use another substitution:

Let (x^{frac{1}{3}} z) implies (x z^3) implies (dx 3z^2dz)

When (x 0), (z 0) and when (x 8), (z 2)

Substitute (z) and (dx) into the original integral:

I (int_{0}^{8} frac{1}{1 x^{frac{1}{3}}}dx int_{0}^{2} frac{1}{1z} 3z^2dz 3int_{0}^{2} frac{z^2}{1z}dz 3int_{0}^{2} frac{z^2 - 1}{1z} dz 3int_{0}^{2} (z - 1) frac{1}{z} dz) 3 ((int_{0}^{2} z - 1 dz frac{z^2}{2} - ln |z|))

3 ((left[frac{2^2}{2} - ln (2)right] - left[frac{0^2}{2} - ln (0)right]) 3 (2 - ln(3)) 3ln(3)

Conclusion

By carefully applying substitution methods and integrating step-by-step, we can evaluate complex definite integrals. The key is to simplify the integrand through appropriate substitutions and correctly handle the bounds of integration. This process not only reinforces the understanding of integration techniques but also enhances problem-solving skills in calculus.

Additional Resources

For further exploration and practice, consider reviewing the following:

Definite Integrals Khan Academy Integral Calculus