# Definite integration problem

Can you show me step by step to solve the definite integral of 1/(sqrt(3+x)), lower bound 5 and upper bound 7?
I'm not that good at calculus, so please explain each step.
Thanks.
God bless y'all

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HallsofIvy
Homework Helper
$$\int_5^7 \frac{dx}{\sqrt{x+ 3}}= \int_5^7(x+3)^{-\frac{1}{2}} dx$$

Now you can make the subsitution u= x+ 3 so that du= dx and when x= 5, u= 8, when x= 6, u= 10:
$$\int_8^{10} u^{-\frac{1}{2}} du$$

Use $$\int u^n du= \frac{1}{n+1}u^{n+1}+ C$$.

HallsofIvy: thx for your help, but what you gave, i think, is indefinite integral, rather than a definite integral. from what i know, because definite integral is a limit, it has to end in a certain number, without a variable. thx again for your help though

To solve for the definite integral, just apply the fundamental theorem of calculus. The hardest part, which HallsofIvy almost completed, was to find the antiderivative.

oh, i see it...sorry for being an idiot :P
thx, both of you guys.

When I apply the Fundamental theorem of calculus in the last step, for LaTeX Code: \\int_8^{10} u^{-\\frac{1}{2}} du , do I use 8 and 10 as my a and b or do I use 5 and 7?

When I apply the Fundamental theorem of calculus in the last step, for LaTeX Code: \\int_8^{10} u^{-\\frac{1}{2}} du , do I use 8 and 10 as my a and b or do I use 5 and 7?
Use 8 and 10

Char. Limit
Gold Member
Well, if you keep the function in u after integrating, you can just evaluate the bounds for u, and you would use 8 and 10. If you switch u to x+3 after integration, you would evaluate the bounds for x, and you would use 5 and 7.