Integrate $\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x$

1. Analyze the question

We are required to evaluate a definite integral of the polynomial function $(2x^2 + x)$ over the interval $[-1, 2]$.

2. Explain the concept

To solve this, we use the Power Rule for Integration, which states:

For a definite integral $\int_{a}^{b} f(x) \, dx$, we find the antiderivative $F(x)$ and then calculate $F(b) - F(a)$.

3. Show the solution

Step 1: Find the indefinite integral

Step 2: Apply the limits from -1 to 2

Step 3: Substitute the upper limit (2)

Convert 2 to a fraction with denominator 3: $2 = \frac{6}{3}$

Step 4: Substitute the lower limit (-1)

Find a common denominator (6):

Step 5: Calculate $F(2) - F(-1)$

Convert $\frac{22}{3}$ to a fraction with denominator 6: $\frac{44}{6}$

Simplify the fraction by dividing by 3:

4. Identify the answer

The result is $7\frac{1}{2}$, which matches option C.

5. Tip

When dealing with definite integrals involving negative limits, be extremely careful with the signs during the subtraction step ($F(b) - F(a)$). A common mistake is to lose a negative sign when subtracting a negative value, which leads to the wrong answer. Always use brackets to keep your signs organized!