Question

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)?

Answer 1

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`