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

Answer:

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#