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Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer.

∫243x1-x2dx

Short Answer

Expert verified

The value of expression is -32ln(5)and the graph is

Step by step solution

01

Step 1. Given information

An expression is given∫243x1-x2dx

02

Step 2. Simplification

Simplify the expression as,

∫243x1-x2dx=3∫24x1-x2dx=3-12lnx2-124=-32ln(5)-32ln(3)+32ln(3)=-32ln(5)

And the graph is

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Most popular questions from this chapter

Without using absolute values, how many definite integrals would we need in order to calculate the absolute area between f(x) = sin x and the x-axis on [-Ï€2,2Ï€]?

Will the absolute area be positive or negative, and why? Will the signed area will be positive or negative, and why?

For each function f and interval [a, b] in Exercises 27–33, use the given approximation method to approximate the signed area between the graph of f and the x-axis on [a, b]. Determine whether each of your approximations is likely to be an over-approximation or an under-approximation of the actual area.

f(x)=x2,[a,b]=[0,3]left sum with

a) n = 3 b) n = 6

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer.

∫01 12exdx

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess and-check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating.

∫x2+2x+22dx

Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A function f for which the signed area between f and the x-axis on [0, 4] is zero, and a different function g for which the absolute area between g and the x-axis on [0, 4] is zero.

(b) A function f whose signed area on [0, 5] is less than its signed area on [0, 3].

(c) A function f whose average value on [−1, 6] is negative while its average rate of change on the same interval is positive.

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