# Calculus math problem help? Evaluate?

Find the area represented by the definite integral.

∫^1 (upper),0(lower) 6 x dx/ SQ(16 + 3x2)

please provide steps so I know how to do it

Relevance

Try a "u" substitution of u = 16+3x^2, so that du = 6x*dx. Then the integrand is just 1/sqrt(u). If you do use this substitution, don't forget to convert back to x before evaluating at your upper and lower limits of x=1 and x=0. (Or, alternatively, convert them to evaluation points in u, which should be u=16 to 19.)

If you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

• If your problem is the integral from low limit x=0 to upper limit x=1 of {(6x) / [√ (16 + 3x²)] } I feel pretty confident that you're going to use a trig substitution [that is, f(x) is going to be a sinΘ, tanΘ, or secΘ -- note that these are the non-"co" functions. In your integral it seems you're going to use a tanΘ. more to come...

Note that your radicand is 4² + [(√3)x]². The cue to use trig sub is when the radicand is [f(x)]² ± [a]²

Let (√3)x/4 = tanΘ [sketch this onto a right triangle, numerator as opp-side and denom-4 adj-side, and follow thru for hypotenuse.] Then x = 4tanΘ/(√3) and dx = [4/(√3)]sec² Θ] dΘ. Meanwhile your radical expression becomes 4sec Θ (because of the proposed substitution). Replacing all your integrand

variable items in terms of Θ you find you're integrating (secΘ) tanΘ dΘ which is NICE. The subsequent integral secΘ can be evaluated by back substitution (from your triangle) or from determining x=0 implies tanΘ = 0; when x=1 tanΘ=(√3)/4

I get a final integral to be ≈ .718 or more precisely 2[(√19) - 4].