Answer the following True or False. For all $1 < a < b$ , $\int_{a}^{b} x^{2} \, dx > \int_{a}^{b} x \, dx$ . $\newline$ True $\newline$ False

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Q. Answer the following True or False. For all

1 < a < b

\int_{a}^{b} x^{2} \, dx > \int_{a}^{b} x \, dx

\newline

True

\newline

False

Calculate Integral of $x^2$ : Let's first calculate the integral of $x^2$ from $a$ to $b$ . The antiderivative of $x^2$ is $(1/3)x^3$ . So, the definite integral from $a$ to $b$ is $(1/3)b^3 - (1/3)a^3$ .
Calculate Integral of $x$ : Now, let's calculate the integral of $x$ from $a$ to $b$ . The antiderivative of $x$ is $(1/2)x^2$ . So, the definite integral from $a$ to $b$ is $(1/2)b^2 - (1/2)a^2$ .
Comparison of Integrals: We need to compare $(\frac{1}{3})b^3 - (\frac{1}{3})a^3$ with $(\frac{1}{2})b^2 - (\frac{1}{2})a^2$ . Since $1 < a < b$ , we know that $b^3 > b^2$ and $a^3 > a^2$ . Therefore, $(\frac{1}{3})b^3 > (\frac{1}{2})b^2$ and $(\frac{1}{3})a^3 > (\frac{1}{2})a^2$ .
Consider Coefficients: However, we need to be careful with the comparison because the coefficients $(\frac{1}{3})$ and $(\frac{1}{2})$ affect the values. $\newline$ We can't directly conclude that $(\frac{1}{3})b^3 - (\frac{1}{3})a^3$ is greater than $(\frac{1}{2})b^2 - (\frac{1}{2})a^2$ just because $b^3 > b^2$ and $a^3 > a^2$ . $\newline$ We need to consider the entire expression.
Analyzing Expressions: Let's analyze the expressions further. $\newline$ For $b^3$ to be greater than $b^2$ , $b$ must be greater than $1$ , which is given. $\newline$ Similarly, for $a^3$ to be greater than $a^2$ , $a$ must be greater than $1$ , which is also given.
Compare Coefficients: Now, let's compare the coefficients. $\newline$ $(\frac{1}{3})$ is less than $(\frac{1}{2})$ , which means that for the same base, the term with the coefficient $(\frac{1}{3})$ will be smaller than the term with the coefficient $(\frac{1}{2})$ . $\newline$ This means that we need to be cautious in our comparison because the larger exponent is paired with the smaller coefficient.
Behavior of Functions: To make a proper comparison, we can consider the behavior of the functions $x^2$ and $x$ as $x$ increases. $\newline$ The function $x^2$ grows faster than the function $x$ as $x$ increases beyond $1$ . $\newline$ This means that the area under the curve of $x^2$ from $a$ to $b$ will grow more rapidly than the area under the curve of $x$ from $a$ to $b$ .
Conclusion: Since the growth rate of $x^2$ is faster than that of $x$ , and both $a$ and $b$ are greater than $1$ , the integral of $x^2$ from $a$ to $b$ will indeed be greater than the integral of $x$ from $a$ to $b$ . This means that the original statement is true.