Solving Integral 7x \sqrt[3]{2-x^2} Dx A Step-by-Step Guide

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Hey everyone! Today, we're diving into the world of integral calculus, specifically focusing on how to solve integrals using a technique called u-substitution. This method is super handy for simplifying complex integrals and making them much easier to tackle. We're going to break down the process step-by-step, using a specific example to illustrate each stage. So, grab your calculators and let's get started!

Understanding the Integral

First off, let's take a look at the integral we're going to solve:

∫7x2−x23dx\int 7x \sqrt[3]{2-x^2} dx

This might look a bit intimidating at first glance, but don't worry! The key to cracking this is recognizing that we can use u-substitution to simplify it. U-substitution, also known as the substitution rule, is essentially the reverse of the chain rule in differentiation. It allows us to replace a complex expression within the integral with a single variable, making the integration process more manageable.

To successfully apply u-substitution, we need to identify a suitable part of the integrand (the function inside the integral) to be our 'u'. This usually involves looking for a function and its derivative (or a multiple thereof) within the integral. In our case, the expression under the cube root, 2-x^2, seems like a good candidate because its derivative will involve x, which we also have in the integrand. This initial step is crucial; selecting the right 'u' can make or break the problem. A wrong choice might lead to a more complicated integral, whereas the right choice simplifies it beautifully.

So, let’s break down the integrand piece by piece. We have 7x, which is a simple polynomial term. We also have the cube root, \sqrt[3]{2-x^2}, which is a composite function. Composite functions are prime candidates for u-substitution because they often contain an inner function and its derivative (or a related term) as part of the integrand. The constant 7 is just a scalar multiple, which we can deal with easily. Recognizing these components helps us formulate a strategy for simplification. Remember, the goal is to transform the integral into a form that we can directly integrate using basic rules. This often involves reducing the complexity of the expression by substituting parts of it with a new variable.

Finding the Change of Variables

The next step is to find a change of variables to rewrite the integrand in terms of u. As we discussed, we'll let:

u=2−x2u = 2 - x^2

This choice is strategic because the derivative of u with respect to x will give us a term involving x, which is present in our original integral. This is the essence of u-substitution – finding a 'u' such that its derivative helps us simplify the integral. The expression 2 - x^2 is inside the cube root, making it a good candidate for u because it complicates the integrand. By substituting this expression, we aim to remove the complexity and make the integral easier to handle.

Now, let's find the derivative of u with respect to x:

dudx=−2x\frac{du}{dx} = -2x

To rewrite the integral in terms of u, we need to express dx in terms of du. We can do this by rearranging the above equation:

du=−2xdxdu = -2x dx

Notice that we have 7x dx in our integral, and we have -2x dx in our expression for du. We can adjust this to match our integral by multiplying both sides by -7/2:

−72du=7xdx-\frac{7}{2} du = 7x dx

This is a crucial step. We've now expressed 7x dx in terms of du, which means we can directly substitute these into our original integral. This substitution will transform our integral from being in terms of x to being in terms of u, which should be a simpler form. The coefficient -7/2 comes from the need to match the 7x in the original integral. We manipulated the du equation to isolate 7x dx so we can replace it directly in the integral. This careful manipulation is what makes u-substitution work.

So, now we have our change of variables:

u=2−x2u = 2 - x^2

du=−2xdx  ⟹  −72du=7xdxdu = -2x dx \implies -\frac{7}{2} du = 7x dx

Rewriting the Integrand

Now that we have our u and du, we can rewrite the integrand in terms of u. Let's go back to our original integral:

∫7x2−x23dx\int 7x \sqrt[3]{2-x^2} dx

We know that u = 2 - x^2, so we can replace 2 - x^2 with u under the cube root:

∫7xu3dx\int 7x \sqrt[3]{u} dx

And we also know that 7x dx = -\frac{7}{2} du, so we can substitute this into the integral as well:

∫−72u3du\int -\frac{7}{2} \sqrt[3]{u} du

We can rewrite the cube root as a fractional exponent:

∫−72u13du\int -\frac{7}{2} u^{\frac{1}{3}} du

Voila! We've successfully rewritten the integrand in terms of u. This form is much simpler to integrate than the original. The u-substitution has transformed a complex integral into a basic power rule integration problem. This step is where the real magic of u-substitution happens. By changing the variable of integration, we've simplified the expression to a form that we can directly apply standard integration rules.

The process of rewriting the integrand is critical. It involves careful substitution and algebraic manipulation to ensure that the integral is entirely in terms of the new variable u. This often requires expressing both the function and the differential (dx) in terms of u and du, respectively. A common mistake is to substitute for the function but forget to change the differential. Remember, the goal is to create an integral that is easier to evaluate, and this rewriting step is the key to achieving that.

Evaluating the Integral

Now, let's evaluate the given integral in terms of u. We have:

∫−72u13du\int -\frac{7}{2} u^{\frac{1}{3}} du

The constant -7/2 can be pulled out of the integral:

−72∫u13du-\frac{7}{2} \int u^{\frac{1}{3}} du

Now we can apply the power rule for integration, which states that:

∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C

where C is the constant of integration. Applying this rule to our integral, we get:

−72⋅u13+113+1+C-\frac{7}{2} \cdot \frac{u^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} + C

Simplify the exponent and the denominator:

−72⋅u4343+C-\frac{7}{2} \cdot \frac{u^{\frac{4}{3}}}{\frac{4}{3}} + C

Divide by the fraction by multiplying by its reciprocal:

−72⋅34u43+C-\frac{7}{2} \cdot \frac{3}{4} u^{\frac{4}{3}} + C

Multiply the fractions:

−218u43+C-\frac{21}{8} u^{\frac{4}{3}} + C

Great! We've integrated the expression in terms of u. However, our original integral was in terms of x, so we need to substitute back to get the final answer in terms of x. This is a vital step in u-substitution. We've done all the calculus, but we need to express the result in the original variable to answer the question completely. Forgetting to substitute back is a common error, so always double-check this step.

The power rule is a fundamental concept in integral calculus, and mastering its application is crucial for solving a wide range of integrals. The constant of integration, C, is added because the derivative of a constant is zero, so there are infinitely many functions that could have the same derivative. Always remember to include C in your indefinite integrals. The steps we've taken so far demonstrate a systematic approach to evaluating integrals using substitution and the power rule.

Substituting Back

Now, let's substitute back for u. We know that u = 2 - x^2, so we replace u in our expression:

−218(2−x2)43+C-\frac{21}{8} (2 - x^2)^{\frac{4}{3}} + C

And that's our final answer! We've successfully evaluated the integral using u-substitution.

This final step is the culmination of all our efforts. We take the result we obtained in terms of u and replace u with its original expression in terms of x. This gives us the solution to the original integral problem. The expression -\frac{21}{8} (2 - x^2)^{\frac{4}{3}} + C represents the family of functions whose derivative is 7x \sqrt[3]{2-x^2}. The constant C accounts for the fact that there are infinitely many antiderivatives, each differing by a constant.

To recap, the process of substituting back involves reversing the initial substitution we made. We replace the temporary variable u with its equivalent expression in terms of the original variable, x. This is crucial for expressing the final result in the context of the original problem. It’s like translating a solution back into the language of the initial question. Without this step, our solution would be in terms of a variable that has no meaning in the original problem. Therefore, always remember to substitute back to complete the solution.

Final Answer

So, to recap, we started with the integral:

∫7x2−x23dx\int 7x \sqrt[3]{2-x^2} dx

We found the change of variables:

u=2−x2u = 2 - x^2

du=−2xdx  ⟹  −72du=7xdxdu = -2x dx \implies -\frac{7}{2} du = 7x dx

We evaluated the integral and substituted back to get the final answer:

−218(2−x2)43+C-\frac{21}{8} (2 - x^2)^{\frac{4}{3}} + C

And that's it! We've successfully navigated the world of u-substitution. Remember, the key is to identify a suitable 'u', find its derivative, rewrite the integral, evaluate, and substitute back. With practice, you'll become a u-substitution pro in no time!

This entire process demonstrates the power and elegance of u-substitution. By strategically changing the variable of integration, we can transform complex integrals into simpler, manageable forms. The technique relies on recognizing composite functions and their derivatives within the integrand. The steps are methodical and logical, and with practice, you can become proficient at applying this method to a wide range of integration problems. Always remember to check your work by differentiating the result to see if you obtain the original integrand. This is a good way to ensure that you have performed the integration correctly.

The beauty of u-substitution lies in its ability to simplify integrals that would otherwise be very difficult or impossible to solve using basic integration rules. It's a technique that showcases the interconnectedness of differentiation and integration, and it's a valuable tool in the arsenal of any calculus student or professional. By mastering u-substitution, you open the door to solving a vast array of integral problems, enhancing your understanding of calculus and its applications in various fields such as physics, engineering, and economics.

Keep practicing, and you'll find that u-substitution becomes second nature. Happy integrating!