4. (5 pts) Let g(x) = ∫2^x (1)/(ln t) dt. (a) Find g'(x)....

4. (5 pts) Let $g(x) = \int_2^x \frac{1}{\ln t} dt$. (a) Find $g'(x)$. Determine if g is increasing or decreasing over its domain $(1, \infty)$. (b) Let $f(x) = \int_{1+x^2}^x \frac{1}{\ln t} dt$. Find $f'(x)$. (c) Use a substitution to show that $g(x) = \int_{\ln 2}^{\ln x} \frac{e^u}{u} du$. (This is a well-known function that has no elementary antiderivative.)

Transcript

00:01 G of x equal to integral f of t d t 0 to x so here in the part 4 we have to evaluate g of 0 and g of 6 so you can see that here g of 0 equal to integral 0 to 0 f of t d t and f of t equal to 0 because integral with equal lower and upper bounds is 0 and g of 6 equal to integral 0 to 6 f of t d t equal to 0 because the graph of f looks like it has rotational symmetric about the point 0 to 3 so the part from x equal to 0 to 3 is equal in area to part from x equal to 3 to 6 but part x equal to 3 to 6 is negative so they will cancel out and we get g of 6 equal to 0 so it is over final answer for part e now we solve part b so here we have to estimate g of x for x equal to 1 2 3 4 and 5 so here we have to count square between the graph of f and the x -axis and you can see that the square below the x -axis are negative so you can see that g of 1 approximate value is 2 .8 for g of 2 the approximate value is 2 .8 plus 1 .9 so here proximate value is 4 .7 now for g of 3 approximate value is 2 .8 plus 1 .9 plus 4 .7 so, 0 .6.

02:53 So now we add them and approximate value is 5 .3.

03:00 Now we have to find the value of g of 4.

03:04 So here approximate value is 2 .8 plus 1 .9 plus 0 .6.

03:13 Here you can see that the square below the x -axis so it will be negative.

03:18 So here negative.

03:19 So here negative 0 .6.

03:22 Now we add that.

03:23 So here proximate value 4 .7 for g of 5 approximate value so here 2 .8 plus 1 .9 plus 0 .6 negative 0 .6 negative 1 .9 so approximate value of g of 5 is 2 .8 so you can see that approximate value of 3 .5 of 3 .8.

03:50 So you can see that approximate value of g of 1 is 2 .8 approximate value of g of 2 .2 is 4 .7 and approximate value of g of 3 is 5 .3 and approximate value of g of 4 is 4 .7 and approximate value of g of 5 5 5 .5 is 2 .8 so it is our final answer for part b now we solve parts so you can see that from x equal to 0 to 3 the area between the curve and the x -axis is all positive then negative after all so you can see that g is increasing on 0 to 3 so it is our final answer for part c now we solve part d so here you can see that from x equal to 0 to 3 the area between the curve and the x -axis is positive then negative afterward so the cumulative area is greatest at x equal to so here g is maximum x equal to 3...

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