User:Xhuwin

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THANK YOU, SHERWIN ~nona

1. Estimate the area under the graph of f(x) = 1/x from x = 1 to x = 3 using four approximating rectangles and right endpoints. Round answer to the nearest whole number.

2. Use the Midpoint Rule with the given value of n to approximate the integral: $\int\limits_{0}^{5}\sin\sqrt{x} dx$ n = 5. Round answer to the nearest whole number.

3. Use the definition of a definite integral to evaluate the integral $\int_{0}^{3}\frac{20}{9} + x - x^2\ dx$ (form: $\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x)$ .

4. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. Then evaluate the derivative when x = 2. $g(x) = \int\limits_{2}^{u}\frac{1}{x^4-x^3}\ dx$

5. Evaluate the integral: $\int_{1}^{2.5} {\frac{2x^2 + t^2\sqrt{x} - 1}{x^2}} dx$ . Round answer to the nearest whole number.

6. Evaluate the integral: $\int_{0}^{.486} {x^{3/7}} dx$ . Round answer to the nearest hundredths.

7. Evaluate the integral: $\int_{0}^{3} {\frac{1}{2x + 3}} dx$ .

8. Evaluate the integral: $\int_{-2}^{3} |\bar{\frac{1}{8}{((x)^2-1)}}| dx$ .

   *  1 recipe pastry for a 9 inch double crust pie
   * 4 tablespoons quick-cooking tapioca
   * 1/8 teaspoon salt
   * 1 cup white sugar
   * 4 cups pitted cherries
   * 1/4 teaspoon almond extract
   * 1/2 teaspoon vanilla extract
   * 1 1/6 tablespoons butter

Solutions:

1. The width of each rectangle is ½. The endpoints are at f (1.5), f (2), f (2.5), f (3). A = ½*[ f (1.5) + f (2) + f (2.5) + f (3)] = 0.95 = 1 (rounded) = 1 9-inch pie crust

2. The width of each rectangle is 1. The midpoints are at f (.5), f (1.5), f (2.5), f (3.5), f (4.5). A = 1*[ f (.5) + f (1.5) + f (2.5) + f (3.5) + f (4.5)] = 4.3978 = 4 (rounded) = 4 tablespoons quick-cooking tapioca

3. $\Delta x = \frac{3}{n} ; x_i = 0 + \frac{4i}{n} ; i = \frac{n(n+1)}{2} ; i^2 = \frac{n(n+1)(n+2)}{6}$

$\lim_{n\to\infty}\sum_{i=1}^n \frac{20}{9}x - x^2\, dx$

$= \frac{20*3i}{9n} - \frac{3^2i^2}{n^2}), \frac{3}{n}$

$= \frac{20*3^2i}{9n^2} - \frac{3^3i^2}{n^3}$

$= \frac{20(n+1)}{2n} - \frac{27(n+1)(2n+1)}{6n^2}$

$= \frac{10n}{n} + \frac{1}{n} - \frac{27n^2}{3n^2} + \frac{27n}{2n^2} + \frac{27}{n^2}$

$= 10 + 0 - 27 / 3 + 0 + 0$

= 1 cup of white sugar

4. $g'(x) = \frac{1}{x^4-x^3} g'(2) = \frac{1}{2^4-2^3} g'(2) =$ 1/8 teaspoon salt

5. $\int_{1}^{2.5} {\frac{2x^2 + t^2\sqrt{x} - 1}{x^2}} dx = \int_{1}^{2.5} {2 + x^{1/2} - x^{-2}} dx = 2x + {2/3}x^{3/2} + \frac{1}{x} = (2*2.5 + {2/3}*2.5^{3/2} + \frac{1}{2.5}) - (2*1 + {2/3}*1^{3/2} + \frac{1}{1}) =$

4 cups pitted cherries

6. $\int_{0}^{.486} {x^{3/7}} dx = \frac{7x^{10/7}}{10} = (\frac{7*.486^{10/7}}{10}) - (\frac{7*0^{10/7}}{10}) = \frac{1}{4}$ teaspoon almond extract

7. $u = 2x + 3, du = 2xdx \int_{3}^{9} \left( \frac{du}{2u} \right) = ( \frac{1}{2}) ln |\bar{u}| ]_3^9 = ( \frac{1}{2}) ln|\bar{9/3}| = ( \frac{1}{2}) ln|\bar{3}|$ teaspoon vanilla extract

8. $\int_{-1}^{-2} |\bar{\frac{1}{8}{((x)^2-1)}}| dx + \int_{-1}^{1} |\bar{\frac{1}{8}{((x)^2-1)}}| dx + \int_{1}^{3} |\bar{\frac{1}{8}{((x)^2-1)}}| dx = \frac{1}{6} + \frac{1}{6} + \frac{5}{6} = \frac{7}{6}$ tablespoons butter

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User:Xhuwin

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