True or False: In many applications of definite integrals, the integral is used to compute the total amount of a varying quantity.

Elleanor Mckenzie
2020-10-28
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hajavaF

Answered 2020-10-29
Author has **90** answers

Definition used:

The integration of the rate of change of a quantity over an interval [a, b] is the result is the total change of that quantity from ato b.

Description:

Application of definite integrals is used to calculate the total amount of the quantity within the range$a\le x\le b$ by the definition mentioned above.

Thus, the integral is used to compute the total amount of a varying quantity in many applications.

Therefore, the given statement is true.

The integration of the rate of change of a quantity over an interval [a, b] is the result is the total change of that quantity from ato b.

Description:

Application of definite integrals is used to calculate the total amount of the quantity within the range

Thus, the integral is used to compute the total amount of a varying quantity in many applications.

Therefore, the given statement is true.

asked 2022-05-01

asked 2021-05-21

Evaluate the integral.

${\int}_{1}^{2}\frac{(x+1{)}^{2}}{x}dx$

asked 2022-04-07

For: $F(0)=0$ and ${F}^{\prime}(x)=f(x)$

Euler's method: $F(0+h)=F(0)+h{F}^{\prime}(0)=0+hf(0)$

Continuing the process, $F(10h)=hf(0)+hf(h)+hf(2h)+.....hf(9h)$

This resembles the Riemann sum: ${\mathrm{\Sigma}}_{i=1}^{n}f({x}_{i})({x}_{i}-{x}_{i-1})$

Therefore my professor used Euler's method to solve integral problems.

Example: ${\int}_{3}^{3.09}f(x)dx=\mathrm{0.81.}$ Approximate $f(3).$ $F(x)=hf(x)$

$0.81=0.09f(x)$

$f(x)=3$

My question: How did $F(x+h)=F(x)+hf(x)$ become $F(x)=hf(x)?$

Euler's method: $F(0+h)=F(0)+h{F}^{\prime}(0)=0+hf(0)$

Continuing the process, $F(10h)=hf(0)+hf(h)+hf(2h)+.....hf(9h)$

This resembles the Riemann sum: ${\mathrm{\Sigma}}_{i=1}^{n}f({x}_{i})({x}_{i}-{x}_{i-1})$

Therefore my professor used Euler's method to solve integral problems.

Example: ${\int}_{3}^{3.09}f(x)dx=\mathrm{0.81.}$ Approximate $f(3).$ $F(x)=hf(x)$

$0.81=0.09f(x)$

$f(x)=3$

My question: How did $F(x+h)=F(x)+hf(x)$ become $F(x)=hf(x)?$

asked 2021-11-07

Find the indefinite integral.

$\int \frac{1}{{x}^{\frac{2}{3}}(1+{x}^{\frac{1}{3}})}dx$

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a) Find the rational zeros and then the other zeros of the polynomial function

b) Factor

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Evaluate $\int {\mathrm{sin}}^{5}xdx$ .

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Answer true or false to the following statement and explain your answer.

A power transformation with an exponent of zero is the logarithm transformation.

A power transformation with an exponent of zero is the logarithm transformation.