1/2 Integral 0 (8x-2x²) dx - 1/2 integral 4 (8x-2x²) dx

To solve the given expression:

\[
\frac{1}{2} \int_0 (8x - 2x^2) \, dx - \frac{1}{2} \int_4 (8x - 2x^2) \, dx
\]

We need to compute both integrals separately.

### Step 1: Compute the first integral:

\[
I_1 = \int_0 (8x - 2x^2) \, dx
\]

First, let's compute the indefinite integral of \(8x - 2x^2\):

\[
\int (8x - 2x^2) \, dx = \int 8x \, dx - \int 2x^2 \, dx
\]

\[
= 4x^2 - \frac{2x^3}{3} + C
\]

Now, evaluate this from 0 to some value (we will substitute the limits later):

\[
I_1 = \left[ 4x^2 - \frac{2x^3}{3} \right]_0^a
\]

Substitute \(x = 0\) and \(x = a\) (for the bounds) to get:

\[
I_1 = \left( 4a^2 - \frac{2a^3}{3} \right) - \left( 4(0)^2 - \frac{2(0)^3}{3} \right)
\]

\[
= 4a^2 - \frac{2a^3}{3}
\]

So, for the first integral:

\[
\frac{1}{2} \int_0 (8x - 2x^2) \, dx = \frac{1}{2} \left( 4a^2 - \frac{2a^3}{3} \right)
\]

### Step 2: Compute the second integral:

\[
I_2 = \int_4 (8x - 2x^2) \, dx
\]

Following the same process, we already know that the indefinite integral is:

\[
\int (8x - 2x^2) \, dx = 4x^2 - \frac{2x^3}{3}
\]

Now, evaluate this from 0 to 4:

\[
I_2 = \left[ 4x^2 - \frac{2x^3}{3} \right]_4^0
\]

\[
I_2 = \left( 4(4)^2 - \frac{2(4)^3}{3} \right) - \left( 4(0)^2 - \frac{2(0)^3}{3} \right)
\]

\[
= \left( 4(16) - \frac{2(64)}{3} \right) - 0
\]

\[
= 64 - \frac{128}{3}
\]

\[
= \frac{192}{3} - \frac{128}{3} = \frac{64}{3}
\]

So, for the second integral:

\[
\frac{1}{2} \int_4 (8x - 2x^2) \, dx = \frac{1}{2} \times \frac{64}{3} = \frac{32}{3}
\]

### Step 3: Combine the results:

Now, substitute the values for \(I_1\) and \(I_2\) into the original expression:

\[
\frac{1}{2} \int_0 (8x - 2x^2) \, dx - \frac{1}{2} \int_4 (8x - 2x^2) \, dx
\]

\[
= \frac{1}{2} \left( 4a^2 - \frac{2a^3}{3} \right) - \frac{32}{3}
\]

Substitute \(a = 0\) (as we have bounds from 0 to 4):

\[
= \frac{1}{2} \left( 4(0)^2 - \frac{2(0)^3}{3} \right) - \frac{32}{3}
\]

\[
= 0 - \frac{32}{3}
\]

\[
= -\frac{32}{3}
\]

Thus, the final result is:

\[
-\frac{32}{3}
\]

---

Do you need further explanation or details on any of the steps?

Here are 5 related questions for further learning:
1. What is the method of solving definite integrals?
2. How can you evaluate integrals with polynomial functions?
3. What happens if the limits of integration are reversed in an integral?
4. How does the constant of integration play a role in indefinite integrals?
5. How can you solve more complex integrals, such as trigonometric or exponential integrals?

**Tip:** Always break down complex integrals into simpler terms, and evaluate step by step, especially when handling polynomial functions!

Learn how to solve the expression \(\frac{1}{2} \int_0 (8x - 2x^2) dx - \frac{1}{2} \int_4 (8x - 2x^2) dx\), which involves definite integrals of polynomial functions. The problem covers key concepts in integral calculus, including definite integrals and polynomial integration. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation