Questions: Find the maximum value of the definite integral I = ∫ from -1 to x (2t^2 - 3t + 1) dt, where x is a real number between 0 and 2, inclusive.

Solution Steps

To find the maximum value of the definite integral \( I = \int_{-1}^{x} (2t^2 - 3t + 1) \, dt \), we first need to compute the integral as a function of \( x \). Then, we find the derivative of this function with respect to \( x \) and set it to zero to find the critical points. Finally, we evaluate the integral at these critical points and the endpoints \( x = 0 \) and \( x = 2 \) to determine the maximum value.

We start by computing the definite integral \( I = \int_{-1}^{x} (2t^2 - 3t + 1) \, dt \). The result of this integral is given by:

\[ I = \frac{2}{3}x^3 - \frac{3}{2}x^2 + x + \frac{19}{6} \]

Next, we find the derivative of the integral with respect to \( x \):

\[ I' = 2x^2 - 3x + 1 \]

To find the critical points, we set the derivative equal to zero:

\[ 2x^2 - 3x + 1 = 0 \]

Solving this quadratic equation yields the critical points:

\[ x = \frac{1}{2}, \quad x = 1 \]

We evaluate the integral \( I \) at the critical points and the endpoints \( x = 0 \) and \( x = 2 \):

• At \( x = 0 \): \[ I(0) = \frac{19}{6} \]

• At \( x = \frac{1}{2} \): \[ I\left(\frac{1}{2}\right) = \frac{27}{8} \]

• At \( x = 1 \): \[ I(1) = \frac{10}{3} \]

• At \( x = 2 \): \[ I(2) = \frac{9}{2} \]

Now we compare the values obtained:

\[ I(0) = \frac{19}{6} \approx 3.1667, \quad I\left(\frac{1}{2}\right) = \frac{27}{8} = 3.375, \quad I(1) = \frac{10}{3} \approx 3.3333, \quad I(2) = \frac{9}{2} = 4.5 \]

The maximum value among these is:

\[ \max\left(\frac{19}{6}, \frac{27}{8}, \frac{10}{3}, \frac{9}{2}\right) = \frac{9}{2} \]

Final Answer