Integrations and Series: Calculus 2 February 10 скачать в хорошем качестве

Integrations and Series: Calculus 2 February 10

In the live stream, we will review homework problems related to computing integrals and series. We will solve more than 80 problems. Please do not forget to like the live stream and subscribe to our channel. Thanks! In this problem 1 , we are asked to find the antiderivative of the function sine of x to the power 3 times cosine of x to the power 4 Let us compute the integral of sine of x to the power 3 times cosine of x to the power 4 . This function is a product of powers of sine and cosine functions. The power of the sine function is 3 which is odd. And the power of the cosine function is 4 which is even. Since the power of the sine function is 3 which is odd, we can rewrite sine to the power 3 of x as the product of sine to power 2 of x and sine of x Note that the square of the sine function is equal to 1 minus the square of the cosine function. So the quantity sine of x to the power 2 can be expressed solely in terms of cosine of x. So we have, sine of x to the power 2 times sine of x is equal to the expression, 1 minus cosine squared of x, to the power 2 times sine of x. Now let us set u equal to cosine of x and d u equal to negative sine of x. The integral of sine of x to the power 3 times cosine of x to the power 4 d x is equal to the integral of cosine of x to the power 4 times the quantity 1 minus cosine squared of x, to the power 1 times sine of x d x. When we substitute cosine of x by u and sine of x dx by negative d u we obtain. The original integral is now equal to negative the integral of u to the power 4 times the quantity 1 minus u squared to the power 2 d u. When we expand the expression, we obtain The integral of u to the power 6 , minus u to the power 4 d u is equal to u to the power 7 over 7 , + , minus u to the power 5 over 5 plus C. Replacing u by cosine of x gives. Consequently, negative the integral of u to the power 4 times the quantity 1 minus u squared to the power 1 d u is equal to cosine of x to the power 7 over 7 , + , minus cosine of x to the power 5 over 5 plus C. Therefore the integral of sine of x to the power 3 times cosine of x to the power 4 d x is equal to cosine of x to the power 7 over 7 , + , minus cosine of x to the power 5 over 5 plus C. This completes the solution. Thanks for watching the video. Please like and subscribe if you have not yet done so. Thanks! In this problem, we will study the convergence of the power series with terms { 1 } over { n } times 6 x + 5 to the power n . We will precisely determine the radius and the interval of convergence of this series. We define the sequence b n as { 1 } over { n } times 6 x + 5 to the power n . In order To determine the interval of convergence of the series, we will use the ratio test. The series converges if the limit of the absolute value of the ratio of the n+1 term to the nth term is less than 1. Now, we have to compute the limit of ratios. It comes that, the limit as n goes to infinity of b n + 1 over b n is equal to the limit as n goes to infinity of 6 x + 5 to the power n + 1 times 1 , over n + 1 , over { 1 } over { n } times 6 x + 5 to the power n The limit of the ratio can be reorganized as the limit as n goes to infinity of the absolute value of 6 x + 5 to the power n + 1 , over 6 x + 5 to the power n, times 1 , over n + 1 , over { 1 } over { n }. This is the same as the limit as n goes to infinity of the absolute value of 6 x + 5 , over 1 , times 1 , over n + 1 , over { 1 } over { n }. We can pull 6 x + 5 , over 1 out of the limit expression because it does not depend on n. Thus, we only need to compute the limit as n goes to infinity of 1 , over n + 1 , over { 1 } over { n }. The limit of the ratio is equal to the absolute value of 6 x + 5 which must be less than 1 for the series to converge. This is because, The limit as n goes to infinity of 1 , over n + 1 , over { 1 } over { n } is equal to 1. This comes from the fact that this quantity is a fraction of polynomials with the same degree, and the same leading coefficients.