Question

Evaluate the integral...

Answer 1

7
S sqrt(t+1) dt
0

Answer 2

You can use u-substitution. Let u=t+1 and du=dt

Answer 3

i need more help than that

Answer 4

What do you need next?

Answer 5

I need to know how to solve this

Answer 6

Then just follow instruction, DO IT!

Answer 7

Follow what @SBurchetter guide you:
Let u=t+1 and du=dt

Answer 8

All right. You can set up the substitutions as\[\int\limits \sqrt{u}du\] You can rewrite the square root as an exponent.\[\int\limits u^{1/2}du\]Now you use the power rule for integration. You add 1 to the exponent and then divide by that value. 1+(1/2)=3/2 So you would now get\[(2/3)u^{3/2}\]Substituting t+1 for u you get\[(2/3)(t+1)^{3/2}|_0^7\]You can now evaluate from 0 to 7.\[(2/3)(7+1)^{3/2}-(2/3)(0+1)^{3/2} \approx 14.41\]

Answer 9

thank you