Question

If $\displaystyle\int_{-1}^{4} f(x)\,dx = 4$ and $\displaystyle\int_{2}^{4} [3 - f(x)]\,dx = 7$, then $\displaystyle\int_{-1}^{2} f(x)\,dx$ equals

• A. $-2$
• B. 3
• C. 5
• D. 8

Hint:

$\displaystyle\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$. Use this to split or combine integral limits when partial values are known.

Answer 1

The Correct Option is C

To solve the problem, we need to find \(\int_{-1}^{2} f(x) \, dx\), given the values of other integrals involving \(f(x)\).

Given:

• \(\int_{-1}^{4} f(x) \, dx = 4\)
• \(\int_{2}^{4} [3 - f(x)] \, dx = 7\)

First, let's evaluate the second integral:

\(\int_{2}^{4} [3 - f(x)] \, dx = \int_{2}^{4} 3 \, dx - \int_{2}^{4} f(x) \, dx\)

We can solve \(\int_{2}^{4} 3 \, dx\) as follows:

\(\int_{2}^{4} 3 \, dx = 3 \times (4 - 2) = 3 \times 2 = 6\)

Substituting back into the equation:

\(6 - \int_{2}^{4} f(x) \, dx = 7\)

Rearranging gives us:

\(\int_{2}^{4} f(x) \, dx = 6 - 7 = -1\)

Now, we use this to find \(\int_{-1}^{2} f(x) \, dx\):

Using the property of definite integrals:

\(\int_{-1}^{4} f(x) \, dx = \int_{-1}^{2} f(x) \, dx + \int_{2}^{4} f(x) \, dx\)

Substitute the known values:

\(4 = \int_{-1}^{2} f(x) \, dx + (-1)\)

Simplifying gives:

\(\int_{-1}^{2} f(x) \, dx = 4 + 1 = 5\)

Hence, the value of \(\int_{-1}^{2} f(x) \, dx\) is 5. Therefore, the correct option is 5.