Question

If $ \int_{n}^{n+1} g(x) \, dx = n^2 $, $ \forall n \in Z $, then the value of $ \int_{-3}^{3} g(x) \, dx $ is

• A. $19$
• B. $28$
• C. $9$
• D. $27$

Hint:

When dealing with definite integrals over a range that can be broken down into intervals where the integral's value is known, split the integral and sum the results.

Answer 1

The Correct Option is A

Step 1: Split the integral into intervals of length 1.
$$ \int_{-3}^{3} g(x) \, dx = \int_{-3}^{-2} g(x) \, dx + \int_{-2}^{-1} g(x) \, dx + \int_{-1}^{0} g(x) \, dx + \int_{0}^{1} g(x) \, dx + \int_{1}^{2} g(x) \, dx + \int_{2}^{3} g(x) \, dx $$ Step 2: Use the given condition $ \int_{n^{n+1} g(x) \, dx = n^2 $.}
$$ \int_{-3}^{-2} g(x) \, dx = (-3)^2 = 9 $$ $$ \int_{-2}^{-1} g(x) \, dx = (-2)^2 = 4 $$ $$ \int_{-1}^{0} g(x) \, dx = (-1)^2 = 1 $$ $$ \int_{0}^{1} g(x) \, dx = (0)^2 = 0 $$ $$ \int_{1}^{2} g(x) \, dx = (1)^2 = 1 $$ $$ \int_{2}^{3} g(x) \, dx = (2)^2 = 4 $$
Step 3: Sum the values of the integrals.
$$ \int_{-3}^{3} g(x) \, dx = 9 + 4 + 1 + 0 + 1 + 4 = 19 $$ Thus, the value of $ \int_{-3}^{3} g(x) \, dx $ is $ \boxed{19} $.