Evaluate the following, int_0^1 (x^e +e^x) dx. I know the anti-derivative of x^e is x^(e+1)/(e+1). How would i solve this,would i use F(b) +F(a)?

1 Answer
Jan 15, 2017

I got e/(e+1) +e -1, however the answer is 1/(e+1) +e -1,

The correct answer is:

int_0^1 (x^e+e^x) \ dx = 1/(e+1)+e - 1

Explanation:

I did F(1)-F(0) and i did anti-derivative of x^e * derivative of x^e which is ex^(e-1)

int_0^1 (x^e+e^x) \ dx = [x^(e+1)/(e+1)+e^x]_0^1
" " = { (1^(e+1)/(e+1)+e^1) - (0^(e+1)/(e+1)+e^0) }
" " = { (1/(e+1)+e) - (0+1) }
" " = 1/(e+1)+e - 1

Where we used:

1^n = 1 AA n in RR \ \ \ \ \ \ \ \ \ \ \ => 1^(e+1)=1
0^n=0 AA n in RR-{0} \ => 0^(e+1)
x^n=1 AA n in RR \ \ \ \ \ \ \ \ \ \ \ => e^0 = 1