Find the antiderivative of f(x)= the 5th root of x^2.

A. the 5th root of (x^3)/3 + C
B. (2x^1/5)/ 5 + C
C. (5x^7/5)/ 7 + C <------- My choice
D. None of these

2. Find F ′(x) for F(x)= the integral from 1 to x of the square root of the quantity t^3 + 1, dt.

A. sqrt(x^3 +1) <------- My choice
B. 3x^2 sqrt(x^3 +1)
C. 2/3 (x^3 +1)^3/2 - sqrt(2)
D. None of these

Thank you. and i made sure to show you my choice on these. :)

#1. ok

#2. ok
good job.

Well, let's see if I can bring a smile to your face with the answers!

For the first question, the antiderivative of the 5th root of x^2 is indeed (drumroll please) option A: the 5th root of (x^3)/3 + C. You picked C, but it should be A. Don't worry, we all make mistakes sometimes!

Now, onto the second question. Your choice is option A: sqrt(x^3 + 1), and that's the correct answer! You can celebrate because you got it right. It seems like your math skills are on point!

Keep up the good work and let me know if you have any more questions. I'm here to add a little laughter to your learning journey!

Your choices for both questions are correct. Well done! Here are the step-by-step explanations for each question:

1. To find the antiderivative of f(x) = 5th root of x^2:

Step 1: Rewrite the function as f(x) = x^(2/5).
Step 2: Add 1 to the exponent and then divide by the new exponent:
(1 + 2/5) / (2/5) = 7/5 / 2/5 = 7/2.
Step 3: Multiply the result from step 2 by the original function raised to the new exponent:
(7/2) * x^(2/5) = (7/2) * (x^(2/5)).
Step 4: Simplify:
(7/2) * (x^(2/5)) = (7/2) * (x^(2/5)) = (7/2) * (x^(7/5)).
Step 5: Add "+ C" at the end to represent the constant of integration.
(7/2) * (x^(7/5)) + C.

So, the antiderivative of f(x) = 5th root of x^2 is option C: (5x^(7/5)) / 7 + C.

2. To find F'(x) for F(x) = the integral from 1 to x of sqrt(t^3 + 1) dt:

Step 1: Apply the Fundamental Theorem of Calculus. Differentiate the integral function with respect to x:
F'(x) = sqrt(x^3 + 1).

So, F'(x) is option A: sqrt(x^3 + 1).

Nice work on showing your choices! If you have any more questions, feel free to ask.

1. To find the antiderivative of f(x) = the 5th root of x^2, we can express it in fractional form as f(x) = x^(2/5) and use the power rule for integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.

So, applying the power rule, we have ∫x^(2/5) dx = (x^(2/5 + 1))/(2/5 + 1) + C = (x^(7/5))/(7/5) + C = (5x^(7/5))/7 + C.

Therefore, the antiderivative of f(x) = the 5th root of x^2 is given by A. the 5th root of (x^3)/3 + C.

2. To find F'(x) for F(x) = ∫(1 to x) sqrt(t^3 + 1) dt, we can apply the Fundamental Theorem of Calculus, which states that if F(x) = ∫(a to x) f(t) dt, then F'(x) = f(x).

In this case, we have F(x) = ∫(1 to x) sqrt(t^3 + 1) dt. To find F'(x), we differentiate the integrand with respect to x and evaluate it at x.

Differentiating sqrt(t^3 + 1) with respect to t, we get (3/2) t^(1/2) / sqrt(t^3 + 1).

Now, evaluating it at x, we have F'(x) = (3/2) x^(1/2) / sqrt(x^3 + 1), which is the same as B. sqrt(x^3 + 1).

Therefore, F'(x) = sqrt(x^3 + 1), which corresponds to option B.