How do I solve the definite integral of 1/(sqrt(3+x)) with bounds 5 and 7?

[Moonflower]

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

 

[HallsofIvy]

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

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

 

[Moonflower]

HallsofIvy: thanks 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. thanks again for your help though

 

[VeeEight]

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.

 

[Moonflower]

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

 

[Moonflower]

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?

 

[Gregg]

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]

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.