Questions: Evaluate the definite integral [ int-1^2left(x^2-5 xright) d x ] using the Fundamental Theorem of Calculus, Part 2. Enter the exact answer. [ int-1^2left(x^2-5 xright) d x= ]

Evaluate the definite integral
[
int-1^2left(x^2-5 xright) d x
]
using the Fundamental Theorem of Calculus, Part 2.

Enter the exact answer.
[
int-1^2left(x^2-5 xright) d x=
]

Transcript text: Evaluate the definite integral \[ \int_{-1}^{2}\left(x^{2}-5 x\right) d x \] using the Fundamental Theorem of Calculus, Part 2. Enter the exact answer. \[ \int_{-1}^{2}\left(x^{2}-5 x\right) d x= \] Number

Solution

Solution Steps

Step 1: Find the Antiderivative

To evaluate the definite integral

\[ \int_{-1}^{2}\left(x^{2}-5 x\right) d x, \]

we first find the antiderivative \( F(x) \) of the integrand \( f(x) = x^{2} - 5x \). The antiderivative is given by:

\[ F(x) = \frac{x^{3}}{3} - \frac{5x^{2}}{2}. \]

Step 2: Evaluate the Antiderivative at the Limits

Next, we apply the Fundamental Theorem of Calculus by evaluating \( F(x) \) at the upper limit \( x = 2 \) and the lower limit \( x = -1 \):

\[ F(2) = \frac{2^{3}}{3} - \frac{5 \cdot 2^{2}}{2} = \frac{8}{3} - \frac{20}{2} = \frac{8}{3} - 10 = \frac{8}{3} - \frac{30}{3} = -\frac{22}{3}, \]

\[ F(-1) = \frac{(-1)^{3}}{3} - \frac{5 \cdot (-1)^{2}}{2} = -\frac{1}{3} - \frac{5}{2} = -\frac{1}{3} - \frac{15}{6} = -\frac{1}{3} - \frac{30}{6} = -\frac{31}{6}. \]

Step 3: Calculate the Definite Integral

Now, we subtract the evaluation at the lower limit from the evaluation at the upper limit:

\[ \int_{-1}^{2}\left(x^{2}-5 x\right) d x = F(2) - F(-1) = -\frac{22}{3} - \left(-\frac{31}{6}\right). \]

To perform this subtraction, we need a common denominator:

\[ -\frac{22}{3} = -\frac{44}{6}, \]

thus,

\[ -\frac{44}{6} + \frac{31}{6} = -\frac{44 - 31}{6} = -\frac{13}{6}. \]