Solve for the product of integral limits
Published:
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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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: $n \cdot m$
Let's start by calculating the definite integral. We'll integrate each term of the quadratic expression one by one.
Step 1: Integration
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.
Now, we apply the Fundamental Theorem of Calculus by substituting the upper limit n and the lower limit m.
Let's group the terms with the same denominators together to make it easier to factorize.
Now we can use the algebraic identities for the difference of cubes and the difference of squares.
Step 2: Factorization
Wait, we know that n minus m is equal to three. Let's substitute this value into our equation.
Substituting three for n minus m in every term, the equation simplifies significantly.
We can cancel out the threes in the first term and simplify the rest.
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