-
Solving Linear Equations II
Problem 253: Solve for in the following equation: Solution: First, we will simplify the fractions that can be simplified. Clear the denominator by multiplying by the least common multiple, 8. Simplify by combining like terms. Solve for . That is, .
-
Solving Linear Equations
Problem 252: Solve the equation . Solution: Let’s “get rid of” the parentheses first. Next, we will combine like terms. Simplify by combining line terms. Solve for .
-
Series – Ratio and Root Tests
Problem 251: Consider the series . Evaluate the following limit. Solution: We will let This means that One can now compute our limit. That is,
-
Radius of Converge of Series
Problem 250: The power series of around is given by (1) Find the radius of convergence. Solution: One can use the ratio test to find the radius of convergence. That is, we an value such that (2) for all . Thus (3) That is, we want (4) which is true for…
-
Series – Taylor and Maclaurin Series
Problem 249: Find the first five non-zero terms of Taylor series centered at for the function below. (1) Solution: The Taylor series of a continuous function is given by (2) where is the point of expansion and are the coefficients. That is, (3) In our case so that (1) takes the form…
-
Applications: Approximation of Area II
Problem 248: The area of the region that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles. (1) Use the above definition to determine the expression that represents the area under the graph of from to , and evaluate the limit. That is,…
-
Applications: Approximation of Area
Problem 247: Consider the integral (1) (a) Find the Riemann sum for this integral using right endpoints and . (b) Find the Riemann sum for this same integral, using left endpoints and . Solution: The Riemann sum is given by (2) where is the number of triangles we use and is the size…
-
Applications: Area Between Curves III
Problem 246: Find the area of the region that is enclosed between and . Solution: In order to find the area enclosed by the two functions, we need to find the limits of integration. That is, since we have two functions equal to , we can make it equal to each other. (1) Let…
-
Applications: Area Between Curves II
Problem 245: Find the area of the region that is enclosed between and . Solution: In order to find the area enclosed by the two functions, we need to find the limits of integration. That is, since we have two functions equal to , we can make it equal to each other. (1) Let…
-
Applications – Area Between Curves I
Problem 244: Find the area of the region that is enclosed between and . Solution: In order to find the area enclosed by the two functions, we need to find the limits of integration. That is, since we have two functions equal to , we can make it equal to each other. (1) Let…
One Problem at a Time
