Solve for the product of integral limits

MathematicsDefinite IntegralsMedium

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8) $\int_{m}^{n} (2x^{2} + 5x + 4) dx = 6$, $n - m = 3$, $n \cdot m = ?$

$\left[ \dfrac{2x^{3}}{3} + \dfrac{5x^{2}}{2} + 4x \right]_{m}^{n}$

$\dfrac{2n^{3}}{3} + \dfrac{5n^{2}}{2} + 4n - \left( \dfrac{2m^{3}}{3} + \dfrac{5m^{2}}{2} + 4m \right) = 6$

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Step by Step Written Solution

1
Step 1

Hi Buket, let's solve this calculus problem together. We are given an integral equation and a relationship between the bounds, and we need to find the product of n and m.

Problem Overview

Given: $\int_m^n (2x^2 + 5x + 4) dx = 6$

$n - m = 3$

Find: $n \cdot m$

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Step 2

Let's start by calculating the definite integral. We'll integrate each term of the quadratic expression one by one.

Step 1: Integration

$$\int_m^n (2x^2 + 5x + 4) dx = 6$$
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Step 3

Using the power rule for integration, two x squared becomes two thirds x cubed, five x becomes five halves x squared, and four becomes four x.

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Step 4

Now, we apply the Fundamental Theorem of Calculus by substituting the upper limit n and the lower limit m.

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Step 5

Let's group the terms with the same denominators together to make it easier to factorize.

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Step 6

Now we can use the algebraic identities for the difference of cubes and the difference of squares.

Step 2: Factorization

$$n^3 - m^3 = (n-m)(n^2 + nm + m^2)$$
$$n^2 - m^2 = (n-m)(n+m)$$
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Step 7

Wait, we know that n minus m is equal to three. Let's substitute this value into our equation.

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Step 8

Substituting three for n minus m in every term, the equation simplifies significantly.

$$\frac{2 \cdot 3(n^2 + nm + m^2)}{3} + \frac{5 \cdot 3(n+m)}{2} + 4(3) = 6$$
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Step 9

We can cancel out the threes in the first term and simplify the rest.

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About This Question

Subject
Mathematics
Topic
Definite Integrals
Difficulty
Medium
Question Type
Open Ended

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