21-44 Evaluate the integral. ∫1^64 (1+√(x))/(√(x)) d x |...

Name: Problem 39, Chapter 5: Calculus Early Transcendentals
Uploaded: 2021-08-14T01:48:25-08:00
Duration: 3 min 33 s
Channel: Kian Manafi
Description: 21-44 Evaluate the integral. ∫1^64 (1+√(x))/(√(x)) d x

Key Concepts

Definite Integral

A definite integral represents the accumulation of a quantity, typically the area under a curve, between specified limits. It involves evaluating the antiderivative at the upper and lower bounds and then taking the difference, as described by the Fundamental Theorem of Calculus.

Integration of Radical Functions

This concept involves techniques for integrating functions that include radicals, such as square roots and cube roots. Converting radical expressions into fractional exponents is a common strategy that simplifies the integration process by allowing the use of standard power rules.

Power Rule for Integration

The power rule is a basic integration method used for functions of the form x^n. It states that the antiderivative of x^n is x^(n+1)/(n+1), provided that n is not equal to -1. This rule is applied after expressing radical functions in terms of fractional exponents.

Linearity of Integration

The linearity property allows the integral of a sum of functions to be split into the sum of the integrals of each function. This simplifies computation, as individual integrals can be evaluated separately and then combined to form the final result.

Transcript

00:01 For this integral, what we're going to want to do is split up the numerator of this function into two different fractions.

00:07 So we're going to have the integral of 1, 264, of 1 divided by x to the 1 half power, which is the same thing as x to the negative 1 half power, and then plus x to the 1 3rd power divided by x to the 1⁄2 power.

00:22 And so whenever we have a fraction of our variable x, where we have x divided by x and the raise to two different powers, we minus the power in the denominator from that in the numerator.

00:35 So we're going to have a power of one -third minus one -half.

00:39 One -third is equal to two -sixth.

00:41 One -half is equal to three -sixth.

00:43 So this is going to be equal to x to the negative one -sixth power.

00:47 So we're going to have x to negative one -half power plus x to the negative -one -sixth power times d -x.

01:00 And so now we just want to split this into the integral of the first.

01:05 Term plus the integral of the second term and both of these are power rule integrals and so we're just going to add one to the power and then divide by the resulting power so we have x the negative one -half power here for this first integral so we're going to add one to that negative one -half and get a positive one -half and then divide by one -half which is the same thing as multiplying by two we're going to do the same thing here for this x to the negative one half power or negative one six power sorry and so we're going to add one we're going to get five sixth and then we're going to divide by five sixth which is the same as dividing or multiplying by the reciprocal six fifths and we're looking from one to 64.

01:54 So when x is equal to 64 we get two times the square root of 64 and the square root of 64 is eight so we're going to have two times eight which is 16 and then plus six times 64 to the 5th, or to the 5th, 6th power...

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