Evaluate the integral d/dx (e^arctanx)dx on the interval [0,1]?

Answer 1

dear u are integrating a derivative. so the answer is
e^(arctan x) 0 through 1
= e^(π/4) - e^0
= e^(π/4) - 1

Answer 2

If I am reading this right, you are taking the integral of the derivative...so in this case just evaluate e^arctan x at 0 and 1.

Answer 3

put in the calculator. I truthfully have Ti-86, so press 2d and hit calc container, then you definately finally end up fnInt (4x+5,x,4,7)=eighty one. I advise you utilize graphing calculator because of the fact i'm taking calc 127 now!

Answer 4

f(x)=(1/144(x^6))-(7/120x^5)
-(1/24x^4)
+(1/6x^3)+(1/2x^2)+x+C

1=0-0-0+0+0+0+c and c=1

As such :

f(x)=(1/144(x^6))-(7/120x^5)
-(1/24x^4)+(1/6x^3)
+(1/2x^2)+x+1

Answer 5

good luck !