Question

Replace the symbol \(*\) by either \(\leq\) or \(\geq\) so that the resulting expressions are correct. Give your reasons. (a) \(\int_{0}^{1} x^{2} d x * \int_{0}^{1} x^{3} d x\) (b) \(\int_{-1}^{1} x^{2} d x * \int_{-1}^{1} x^{3} d x\) (c) \(\int_{1}^{3} x^{2} d x * \int_{1}^{3} x^{3} d x\).

Short answer

Question: In each of the expressions provided, replace the * symbol by either ≤ or ≥ so that the expression is correct: (a) ∫₀¹ x² dx * ∫₀¹ x³ dx (b) ∫₋₁¹ x² dx * ∫₋₁¹ x³ dx (c) ∫₁³ x² dx * ∫₁³ x³ dx Answer: (a) ∫₀¹ x² dx ≤ ∫₀¹ x³ dx (b) ∫₋₁¹ x² dx ≥ ∫₋₁¹ x³ dx (c) ∫₁³ x² dx ≤ ∫₁³ x³ dx

Step-by-step solution

Identify Positivity/Negativity of Functions

In this case, we have \(x^{2}\) and \(x^{3}\) over the interval \([0,1]\). Both functions are positive in this interval.

Compare orders of functions

We notice that both functions are monomials with different exponents. In the range \([0,1]\), a function with a higher order (exponent) will have higher values for \(x \in (0,1)\). Hence, \(x^{2} < x^{3}\) for \(x \in (0,1)\).

Analyzing the integrals

So, for each \(x\) within the interval \([0,1]\), we have \(x^{2} < x^{3}\). Since both functions are positive on that interval, integrating both functions over the same interval implies: \(\int_{0}^{1} x^{2} d x < \int_{0}^{1} x^{3} d x\). Thus, the correct expression is: \(\int_{0}^{1} x^{2} d x \leq \int_{0}^{1} x^{3} d x\) (b) \(\int_{-1}^{1} x^{2} d x * \int_{-1}^{1} x^{3} d x\)

Identify Positivity/Negativity of Functions

In this interval, \([-1,1]\), the function \(x^{2}\) is always positive while the function \(x^{3}\) is negative when \(x < 0\) and positive when \(x > 0\).

Analyzing the integrals

The integration of the function \(x^{3}\) over the interval \([-1,1]\) is equal to zero, since the area under the curve is symmetric around the origin and since it is negative for \(x < 0\) and positive for \(x > 0\). In other words, the positive and negative areas cancel each other out. Hence, \(\int_{-1}^{1} x^{3} d x = 0\). The integration of the function \(x^{2}\), being always positive, will be greater than 0. Thus, the correct expression is: \(\int_{-1}^{1} x^{2} d x \geq \int_{-1}^{1} x^{3} d x\) (c) \(\int_{1}^{3} x^{2} d x * \int_{1}^{3} x^{3} d x\)

Identify Positivity/Negativity of Functions

In the interval \([1,3]\), both the functions \(x^{2}\) and \(x^{3}\) are positive.

Compare orders of functions

Similar to Step 2 for case (a), in the range \([1,3]\), a function with a higher order (exponent) will have higher values for \(x \in (1,3)\). Hence, \(x^{2} < x^{3}\) for \(x \in (1,3)\).

Analyzing the integrals

So, for each \(x\) within the interval \([1,3]\), we have \(x^{2} < x^{3}\). Since both functions are positive on that interval, integrating both functions over the same interval implies: \(\int_{1}^{3} x^{2} d x < \int_{1}^{3} x^{3} d x\). Thus, the correct expression is: \(\int_{1}^{3} x^{2} d x \leq \int_{1}^{3} x^{3} d x\)