https://file.aiquickdraw.com/mathbot/file/17260433590212b3h7k3a.jpg

The expression in the image consists of three integrals:

\[
\int_{-2}^{0} (3x+2) \, dx - \int_{-2}^{5} (3x+2) \, dx + \int_{0}^{5} (3x+2) \, dx
\]

Let me compute each integral step by step and simplify the entire expression.

### First integral:
\[
\int_{-2}^{0} (3x + 2) \, dx
\]
The integral of \(3x + 2\) is:
\[
\frac{3x^2}{2} + 2x
\]
Evaluating it from \(-2\) to \(0\):
\[
\left(\frac{3(0)^2}{2} + 2(0)\right) - \left(\frac{3(-2)^2}{2} + 2(-2)\right)
= 0 - \left(\frac{3(4)}{2} - 4\right) = 0 - \left(6 - 4\right) = -2
\]

### Second integral:
\[
\int_{-2}^{5} (3x + 2) \, dx
\]
Using the same antiderivative \(\frac{3x^2}{2} + 2x\), evaluate it from \(-2\) to \(5\):
\[
\left(\frac{3(5)^2}{2} + 2(5)\right) - \left(\frac{3(-2)^2}{2} + 2(-2)\right)
= \left(\frac{75}{2} + 10\right) - \left(6 - 4\right)
= \left(\frac{75}{2} + 10\right) - 2 = \frac{75}{2} + 8 = \frac{91}{2}
\]

### Third integral:
\[
\int_{0}^{5} (3x + 2) \, dx
\]
Using the same antiderivative \(\frac{3x^2}{2} + 2x\), evaluate it from \(0\) to \(5\):
\[
\left(\frac{3(5)^2}{2} + 2(5)\right) - \left(\frac{3(0)^2}{2} + 2(0)\right)
= \left(\frac{75}{2} + 10\right) - 0 = \frac{75}{2} + 10 = \frac{95}{2}
\]

### Putting everything together:
The overall expression is:
\[
-2 - \frac{91}{2} + \frac{95}{2}
\]

Simplifying:
\[
-2 - \frac{91}{2} + \frac{95}{2} = -2 + \frac{95 - 91}{2} = -2 + \frac{4}{2} = -2 + 2 = 0
\]

Thus, the value of the entire expression is \(0\).

Would you like a more detailed explanation or have any further questions?

Here are 5 related questions:
1. How do we handle integrals with linear functions like \(3x + 2\)?
2. What is the geometric interpretation of the definite integral?
3. How does the Fundamental Theorem of Calculus apply to evaluating definite integrals?
4. Can we split or combine definite integrals over adjacent intervals?
5. How do we approach integrals of higher-order polynomials?

**Tip:** When working with definite integrals, always carefully evaluate the boundaries and simplify the algebra to avoid small mistakes!

Evaluate the expression consisting of three integrals: ∫(-2 to 0)(3x+2)dx - ∫(-2 to 5)(3x+2)dx + ∫(0 to 5)(3x+2)dx.

Solve the problem by evaluating three definite integrals of the function (3x+2) over different intervals. This solution applies the Fundamental Theorem of Calculus and simplifies the result. Suitable for students in Grades 11-12 and early college.

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