I need to find the definite integral of sin t with respect to t from lower limit 3 to upper limit ln x!!! Help plz
Can anyone help me plz
Shonie, please be sure you integrate sin t with respect to t BEFORE substituting the upper and lower limits of integration. If f(t) = sin t, F(t) (the integral of sin t) is what?
So can u help me or do u just want to correct me and then have someone else answer???? B/c wording it correctly is great but I need help
Shonie: I'm sorry you're frustrated, but I'm trying to guide you through the solution of this problem, and am more interested in your gaining understanding of concepts than I am in your solving just one problem. To return to our previous discussion: If f(t) = sin t, then the integral is F(t) = - cos t. According to the Fundamental Theorem, the integral in question is F(lin x) - F(3). Siimplify the result, and that will be your final answer.
I'm sorry but yes I am frustrated.. So it would be =sin ln x - sin 3
Shonie: First, integrate the function sin t with respect to t. This is equivalent to finding the function F(t) (the integral of f(t) ). Result: F(t) = - cos t. Now evaluate F(ln x) - f(3): -cos(ln x) - cos 3. That's the answer. Post another question or two from the same section in your book or homework right here, and I'll then give you quick feedback on your work. Good luck!
Apologies. I made a sign error in my last post. Correct answer is -cos(ln x) + cos(3). Let me know how you're doing.
\[\int\limits_{3}^{\ln x} dt~\sin t \] this is your integral?
Yes but the dx sin t looks like sin t dt
But this from the fundamental theorem of calculus 1 and it is taking the derivative of the integral function so I have been told from a teacher that the answer is what I had originally = sin ln x* 1/x
Shonie: This last statement of yours is the key to our misunderstanding. Your original presentation of the problem was "I need to find the definite integral of sin t with respect to t from lower limit 3 to upper limit ln x." Now you explain, "But this from the fundamental theorem of calculus 1 and it is taking the derivative of the integral function." I must apologize and take responsibility for not having actually looked up the Fundamental Theorem of Calculus I, which would have jogged my memory and made me aware that you and I were to FIND THE DERIVATIVE OF the function defined as the integral from 3 to ln x of sin t with respect to t.
I understand. I have been studying this all again and have a better understanding of it when I asked last nite I had just learned the concept and was very confused
There are two versions of this Fundamental Theorem of Integral Calculus I: In the first case we seek to find the derivative the function defined as the integral of f(t) with respect to t with lower limit of integration a and upper limit of integration x. According to the Fund. Thm., all we have to do is to take the integrand (which is f(t) ) and replace t with x. Done. The answer is simply f(x). In the calculus problem you've shared, the integrand is sin t, so replacing t with x results in the final "answer": sin x. In the second case, we happen to have as upper limit of integration a function of x. The process of evaluating the derivative of this function defined as an integral is quite similar to that in Case I, above, except that we replace the variable "t" in the integrand with f(x) and then multiply this result with the derivative of f(x). Applying this information to the problem you've presented: "Find the derivative of the function defined as the integral from 3 to ln x of sin t with respect to t." The integrand is sin t. Replace this t with (ln x), obtaining sin(ln x). Finally, multiply this result by the derivative of ln x, which is 1/x. Thus, like your teacher said, the correct answer is sin(ln x)*(1/x). Best wishes.
Thank you! You helped alot
My great pleasure. I'm so glad this collaboration of ours has led to a happy ending!
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