Question

How do you integrate `\int _ { 1} ^ { 16} \frac { t + 1} { \sqrt { t } } d t`?

Answer 1

`int_1^16(t+1)/sqrttdt=22`

Explanation:

Let `sqrtt=x` and then `x^2=t` .

Hence`1/(2sqrtt)dt=dx` or `(dt)/sqrtt=2dx` and

`int_1^16(t+1)/sqrttdt=int_1^4(x^2+1)dx`

= `[x^3/3+x]_1^4`

= `[4^3/3+4-1^3/3-1]`

= `[64/3+4-1/3-1]`

= `63/3+3`

= `22`