/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Evaluate the integrals. $$\int... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals. $$\int(x+1)^{2}(1-x)^{5} d x$$

Short Answer

Expert verified
The integral is \(-\frac{2}{3} (1-x)^6 + \frac{4}{7} (1-x)^7 - \frac{1}{8} (1-x)^8 + C\).

Step by step solution

01

Choose a Substitution

To solve the integral \( \int(x+1)^{2}(1-x)^{5} d x \), we can use substitution. Let \( u = 1-x \). Then, the differential \( du = -dx \) or equivalently \( dx = -du \). Also, we adjust \( x \) in terms of \( u \): \( x = 1-u \).
02

Express the Integral in Terms of u

Substitute \( x = 1-u \) and \( dx = -du \) into the integral: \[ \int (x+1)^2 (1-x)^5 dx = \int ((1-u)+1)^2 u^5 (-du) \].Simplify it to: \[ -\int (2-u)^2 u^5 du \].
03

Expand the Integrand

Expand \( (2-u)^2 \) to simplify the integral: \[ (2-u)^2 = 4 - 4u + u^2 \].Substitute back into the integral: \[ -\int (4 - 4u + u^2) u^5 du = -\int (4u^5 - 4u^6 + u^7) du \].
04

Integrate Term by Term

Integrate each term separately: \[ -\int 4u^5 du + \int 4u^6 du - \int u^7 du \].This becomes: \[ -\left( \frac{4u^6}{6} \right) + \left( \frac{4u^7}{7} \right) - \left( \frac{u^8}{8} \right) + C \].
05

Simplify the Integrated Expression

Simplify the coefficients of the integrated terms: \[ -\frac{2}{3} u^6 + \frac{4}{7} u^7 - \frac{1}{8} u^8 + C \].
06

Substitute Back to x

Recall that \( u = 1-x \). Substitute back \( u = 1-x \) into the expression: \[ -\frac{2}{3} (1-x)^6 + \frac{4}{7} (1-x)^7 - \frac{1}{8} (1-x)^8 + C \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is an essential field of mathematics focused on the operation of integration, which is the process of finding the integral of a function. This process is fundamental in calculating areas under curves and understanding the accumulation of quantities. When dealing with integral calculus, there are two main types of integrals: indefinite and definite integrals.

In the case of indefinite integrals, we are concerned with finding the antiderivative of a function, and the result includes a constant of integration, commonly denoted as \( C \). This constant is crucial because it accounts for the fact that there are infinitely many antiderivatives for any given function, each differing by a constant value.

In solving the given problem using substitution, we focus on indefinite integrals, looking for an expression whose derivative matches our integrand. This technique, called substitution, is a method to simplify complex integrals by replacing variables with expressions that mean the same thing in differential form.
Definite Integrals
Definite integrals are a type of integral that compute the accumulation of a quantity between two distinct points, providing a specific numerical value. They are represented with bounds indicating the interval over which the integration is performed. These bounds essentially serve as the limits of integration. However, in this exercise, we are working with an indefinite integral, which differs from definite integrals because it does not have specified bounds.

While the original problem and solution dealt with an indefinite integral, understanding definite integrals is essential for tackling problems involving real-world applications, like calculating areas, volumes, and other quantities. Unlike indefinite integrals that result in a function plus a constant \( C \), definite integrals directly evaluate to a number, demonstrating the net accumulation of a quantity over the specified range. This translates, in a geometric sense, to the signed area under the curve represented by the function, bounded by the specified limits.
Antiderivatives
Antiderivatives play a fundamental role in integral calculus, as they represent the reverse process of differentiation. The task of finding an integral of a function equates to finding its antiderivative. This serves as a bridge between differential calculus and integral calculus, providing insights into how quantities accumulate over time.

In the problem at hand, we use substitution to simplify the integrand into a form where finding the antiderivative becomes more straightforward. Once the function is expressed in a simplified form \( (-\int (4u^5 - 4u^6 + u^7) \, du) \), we proceed by integrating each term separately, a typical characteristic of handling polynomials. Finding the antiderivatives term by term, followed by substituting back with the original variables, ensures that our solution is representative of the original function. This step highlights the importance of being comfortable with finding antiderivatives for a wide variety of functions.

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Most popular questions from this chapter

Archimedes' area formula for parabolic arches Archimedes (287-212 B.C.), inventor, military engineer, physicist, and the greatest mathematician of classical times in the Western world, discovered that the area under a parabolic arch is two-thirds the base times the height. Sketch the parabolic arch \(y=h-\left(4 h / b^{2}\right) x^{2}\) \(-b / 2 \leq x \leq b / 2,\) assuming that \(h\) and \(b\) are positive. Then use calculus to find the area of the region enclosed between the arch and the \(x\) -axis.

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=\sin ^{2} x \quad \text { on } \quad[0, \pi]$$

Find the area of the region in the first quadrant bounded on the left by the \(y\) -axis, below by the curve \(x=2 \sqrt{y}\), above left by the curve \(x=(y-1)^{2},\) and above right by the line \(x=3-y\).

Graph the function and find its average value over the given interval. $$f(t)=t^{2}-t \quad \text { on } \quad[-2,1]$$

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100,200,\) and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=x \sin ^{2} \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right]$$

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