Mathematics 136 – Calculus 2 Exam 3 – Sample Exam Questions – Solutions

April 15, 2008

I. (A) The arclength is

L = ∫ \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt} )^{2}} dt
= ∫_{0}^{2} \sqrt{36t^{2} + 36t^{4}} dt
= ∫_{0}^{2} 6t\sqrt{1 + t^{2}} dt, so letu = 1 + t^{2}
= 2(1 + t^{2})^{3/2}|_{0}^{2}
= 2(5\sqrt{5} − 1).

(B) The arclength is

L = ∫ \sqrt{1 + (\frac{dy}{dx})^{2}} dx

= ∫_{0}^{3} \sqrt{1 + (\frac{1}{4}(x^{2} + 4)^{1/2}(2x))^{2}} dx

= ∫_{0}^{3} \sqrt{\frac{1}{4}x^{4} + x^{2} + 1} dxa perfect square under the radical
= ∫_{0}^{3} \frac{1}{2}(x^{2} + 2) dx
= \frac{x^{3}}{6} + x|_{0}^{3}
= \frac{15}{2}.

II. (A) The average value is

f_{ave} = \frac{1}{2} ∫_{0}^{2} x\sqrt{1 + x^{4}} dx

To evaluate the integral, use the substitution u = x^{2} and #21 in the table (or a tangent substitution):

f_{ave} = \frac{1}{4} ∫_{0}^{4} \sqrt{1 + u^{2}} du
= \frac{1}{4} \frac{u}{2}\sqrt{1 + u^{2}} + \frac{1}{2} ln(u + \sqrt{1 + u^{2}})|_{0}^{4}
= \frac{1}{2}\sqrt{17} + \frac{1}{8} ln(4 + \sqrt{17}).