University:
Missouri State University
Course:
MTH 136 | Precalculus 1
Academic year:

2019
Views:

21

Pages:

8
Author:

Elisa Morris

Calculus Name: Key Unit 5 Review Date: Block: 1.) Find (\frac{d}{dx} \int_{2}^{x} (1-t^3) dt) 2.) (\frac{d}{dx} \int_{2}^{x^2} (1-t^3) dt) \(-(1-x^3)\) \(2x(1-x^6)\) 3.) Find the interval on which (f(x) = \int_{0}^{x} (t^3 + t^2 + 1) dt) is concave up. (f''(x) = x^3 + x^2 + 1) (f''(x) = 3x^2 + 2x) (3x^2 + 2x = 0) (x(3x + 2) = 0) (x = 0) (x = -2/3) ((-\infty, -2/3) \cup (0, \infty)) 4.) Evaluate using Riemann Sums (L_4) and (R_4) for the following on the interval (0 \le x \le 8) with four subintervals. \(\frac{8-0}{4} = 2 = \Delta x\) [Graph for L4] \(L_4 = 2[0 + 2 + 2.75 + 2.75]\) \(2[7.5] = 15\) [Graph for R4] \(R_4 = 2[2 + 2.75 + 2.75 + 2]\) \(2[9.5] = 19\) 5.) Approximate (\int_{2}^{4} (3x^2 - 2x) dx) using (\Delta x = \frac{4-2}{4} = \frac{1}{2}) a.) Midpoint rule with 4 subintervals b.) Trapezoid rule with 4 subintervals \(\Delta x = \frac{1}{2}\) \(M_4 = \frac{1}{2} [f(2.25) + f(2.75) + f(3.25) + f(3.75)]\) \(\frac{1}{2} [10.188 + 17.188 + 25.188 + 34.888]\) \(\frac{1}{2} [87.752] = 43.876\) [Table for Trapezoid Rule] \(x\) | \(f(x)\) ---|--- 2 | 8 2.5 | 13.75 3 | 21 3.5 | 29.75 4 | 40 \(\frac{1}{2} (0.5) [8 + 2(13.75) + 2(21) + 2(29.75) + 40]\) \(\frac{1}{4} [8 + 27.5 + 42 + 59.5 + 40]\) \(\frac{1}{4} [177] = 44.25\)