Integrating (3-x)7^[(3-x)2]: A Step-by-Step Guide

[Blade]

Need a kickstart with this one:
f(t)=t^(3/2)log(of 2)Sqrt(t+1)

Integral of (3-x)7^[(3-x)^2] dx

7^[(3-x)2] = e^[(3-x)2ln 7]
u=(3-x)^2
du/dx = -2(3-x)
(3-x)dx = -1/2 du
not even sure what so far is right..

 

[HallsofIvy]

I'm not sure what the question is!

"f(t)=t^(3/2)log(of 2)Sqrt(t+1)"

Okay, what's the question?


"Integral of (3-x)7^[(3-x)^2] dx

7^[(3-x)2] = e^[(3-x)2ln 7]
u=(3-x)^2
du/dx = -2(3-x)
(3-x)dx = -1/2 du
not even sure what so far is right.."

Seeing the exponent (3-x)² and (3-x) multiplying the exponential, the first thing I would try is "let u= (3-x)²". Then du= -2(3-x)dx so the integral becomes

-2 times Integral of 7^udu.

If you don't know the derivative and anti-derivative of 7^u, remember that 7^u= e^{u ln(7)}.

 

[Blade]

Ah, sorry about the first one.

f(t)=t^(3/2)log(of 2)Sqrt(t+1)
I need to derive that.

 

[himanshu121]

I feel u want derivative , if so then hint is

take log on both sides and then differentiate

 

[HallsofIvy]

I don't see any reason to take the logarithm. It's looks like a pretty direct application of the product rule and chain rule.

f(t)=t^3/2(log<sub>2[/sup](&radic;(t+1))

f'= (t^3/2)'log2[/sup](&radic;(t+1))+(t²)(log2[/sup](&radic;(t+1))'

(t^3/2)'= (3/2)t^1/2, of course.

To differentiate log2(x) recall that log2(x)= ln(x)/ln(2) so (log2(x))'= 1/(xln(2)).</sub>

 

[himanshu121]

There are many ways of doing a problem, though both are easy to use.

yes it is a direct problem involving the product rule and chain rule