integral upper bound 5 lower bound -2 (g(x)) dx = 8, than integral upper bound 5 lower bound -2 (g(x) + 6)dx = ?

Given the integral:

\[
\int_{-2}^{5} g(x) \, dx = 8
\]

We need to find the value of the integral:

\[
\int_{-2}^{5} (g(x) + 6) \, dx
\]

We can separate the integral into two parts:

\[
\int_{-2}^{5} (g(x) + 6) \, dx = \int_{-2}^{5} g(x) \, dx + \int_{-2}^{5} 6 \, dx
\]

We already know that:

\[
\int_{-2}^{5} g(x) \, dx = 8
\]

Now, let's calculate the second integral:

\[
\int_{-2}^{5} 6 \, dx
\]

Since \( 6 \) is a constant, the integral simplifies to:

\[
6 \int_{-2}^{5} 1 \, dx = 6 \times \left(5 - (-2)\right) = 6 \times 7 = 42
\]

So, adding the results together:

\[
\int_{-2}^{5} (g(x) + 6) \, dx = 8 + 42 = 50
\]

The value of the integral is **50**.

Would you like more details or have any questions?

Here are some related questions:

1. How do you find the integral of a constant function over an interval?
2. What properties of integrals allow us to split integrals into sums?
3. How does adding a constant to a function affect its integral?
4. Can you generalize this process to any interval and function?
5. What happens if the constant is negative instead of positive?

**Tip:** When integrating a sum of functions, you can always split the integral into the sum of integrals for easier calculation.

Learn how to evaluate the definite integral of the function g(x) over the interval from -2 to 5, including the effect of adding a constant. This problem demonstrates the use of integral properties and is 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