# 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

### 2 Answers

- 8 years agoBest Answer
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.)

You can check your answer at the Mathematica website below.

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.

Source(s): http://integrals.wolfram.com/index.jsp - answerINGLv 68 years ago
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].