Evaluate the indefinite integral.∫x(5+x)79dx\int_{}^{}x\left(5+x\right)^{79}dx∫x(5+x)79dx

1 81 ( 5 + x ) 81 − 1 16 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{16}\left(5+x\right)^{80}+C

Evaluate the indefinite integral. ∫ x ( 5 + x ) 79 d x \int_{}^{}x\left(5+x\right)^{79}dx ∫ x ( 5 + x ) 79 d x

this problem, we want to evaluate the indefinite integral of x times five plus x to the power of 79 d x. Now we're going to go ahead and get started here with our first step for substitution, which is choosing u and finding du. Now here we can see this function five plus x enclosed in those parentheses and raised to a power here, so that's a good clue that that's what we should choose to be u, that five plus x. If u is equal to five plus x, that makes du equal to just d x because we can look at the derivative of this. It's just one because the derivative of five is zero. The derivative of x is one. And then multiplying one times d x is just d x. So with u and du found, let's go ahead and move on to step two, rewriting our integral only in terms of this u and du. Now looking at my integral, I can see that I have this u that will be raised to the power of seventy seventy nine and this d x which represents du as we figured out in step one, but that still leaves us with this extra x here. So what can we do about that? Well remember that we can take this and rewrite our x in terms of u so that everything in our integral is in terms of u and du. In order to do that, I can go ahead and just rearrange this equation by subtracting five from each side. If I subtract five from each side here, that will cancel out over here and leave me with x being equal to u minus five. So that means that this x being a u minus five, I can go ahead and rewrite this integral as the integral of u minus five times u to the power of 79 du. This is all in terms of u and du and we can continue on to step three. I'm gonna go ahead and distribute this u to the power of 79 into my integral just to make it a little bit easier to deal with here and I'm gonna go ahead and move this x down here just so that I have a little bit more space to work with. So going ahead and distributing that u to the power of 79, that makes this a u to the power of 80 because this is u to the one, just adding those exponents together and then minus five times u to the power of 79 du. Now we can go ahead and just apply the power rule here when finding this antiderivative. So decreasing, increasing that power by one and dividing by that new exponent, so here 80 plus one. So that first term is going to be one over 81 times u to the power of 81 and then for this next one minus five, keeping that constant out here for now, and then increasing that power by a one dividing up by that new power, that's going to be one over 80 times u to the power of 80 and then plus that constant of integration c. Now we can do some simplification here because this five over 80 as a fraction, I can reduce that to be one over 16. So I have one over 81 times u to the power of a one minus one over 16 having reduced that fraction, u to the power of 80 plus c. Then my final step here is to replace u with that original function of x. In this case, that was five plus x. So I want to replace both of my u's here with that. That's gonna be one over 81 times five plus x to the power of 81 minus one over 16 times five plus x to the power of 80 plus my constant of integration is c. And this is my final answer here. Feel free to let us know if you have any questions here and let's keep practicing!

Evaluate the indefinite integral. ∫ x ( 5 + x ) 79 d x \int_{}^{}x\left(5+x\right)^{79}dx ∫ x ( 5 + x ) 79 d x

1 81 ( 5 + x ) 81 − 1 16 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{16}\left(5+x\right)^{80}+C

1 81 x 81 − 1 16 x 80 + C \frac{1}{81}x_{}^{81}-\frac{1}{16}x^{80}+C

1 81 ( 5 + x ) 81 − 1 80 ( 5 + x ) 80 + C \frac{1}{81}\left(5+x\right)^{81}-\frac{1}{80}\left(5+x\right)^{80}+C

1 81 x 81 − 1 80 x 80 + C \frac{1}{81}x^{81}-\frac{1}{80}x^{80}+C

8. Definite Integrals

Substitution for Indefinite Integrals

Business Calculus

8. Definite Integrals

Substitution for Indefinite Integrals

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}x\left(5+x\right)^{79}dx$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{16} {(5 + x)}^{80} + C$

$\frac{1}{81} x_{}^{81} - \frac{1}{16} x^{80} + C$

$\frac{1}{81} {(5 + x)}^{81} - \frac{1}{80} {(5 + x)}^{80} + C$

$\frac{1}{81} x^{81} - \frac{1}{80} x^{80} + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a product of x and a power of (5 + x). To simplify, use the substitution method. Let u = 5 + x, which implies that du/dx = 1 or du = dx.

Step 2: Rewrite x in terms of u. Since u = 5 + x, we can express x as u - 5. Substitute this into the integral, replacing x with (u - 5) and dx with du.

Step 3: The integral now becomes ∫(u - 5)u^79 du. Expand the integrand by distributing u^79 to both terms, resulting in ∫(u^80 - 5u^79) du.

Step 4: Integrate each term separately. For ∫u^80 du, use the power rule for integration: ∫u^n du = (u^(n+1))/(n+1). Similarly, for ∫5u^79 du, factor out the constant 5 and apply the power rule.

Step 5: After integrating, substitute back u = 5 + x to return to the original variable. Add the constant of integration C to complete the indefinite integral.

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