# consider the function

f(x) = (x if x<1
(1/x if x>or equal to 1

Evaluate the definite integral:

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

## You have to say what is the lower limit and what is the upper limit of your integral to do a definite integral

f(x) = x for x < 1
and
f(x) = 1/x for x >/= 1

(-2)^3 = -8

so when x<1
integral = -8 [x^2/2 upper-x^2/2 lower]
and for x >/= 1
integral = -8 [ln x upper - ln x lower]

## ∫[-2,3] f(x) dx

= ∫[-2,1] x dx + ∫[1,3] 1/x dx

Now you can take it from there.

Note that f(1) is not defined.

## Ah, evaluating definite integrals, an activity that warms my clown heart. Let's dive right in, shall we?

We need to split the integral into two parts since the function changes at x = 1.

For x < 1, we integrate f(x) = x from -2 to 1:
∫_{-2}^{1} x dx

Using the power rule of integration, we get:
[1/2 * x^2]_{-2}^{1}

Plugging in the values, we have:
[1/2 * (1)^2] - [1/2 * (-2)^2]

Simplifying, we get:
1/2 - 2 = -3/2

Now for x ≥ 1, we integrate f(x) = 1/x from 1 to 3:
∫_{1}^{3} (1/x) dx

Using the logarithmic rule of integration, we get:
[ln| x |]_{1}^{3}

Plugging in the values, we have:
ln|3| - ln|1|

Simplifying, we get:
ln(3) - ln(1) = ln(3)

Finally, to find the definite integral of f(x), we add the two parts together:
int_{-2}^{3} f(x) dx = -3/2 + ln(3)

And there you have it! The result of the definite integral is -3/2 + ln(3). I hope that brought a smile to your face!

## To evaluate the definite integral of the function $$f(x)$$, we need to split the integral into two parts based on the interval of $$x$$.

1. For $$x < 1$$: In this case, the function is simply $$f(x) = x$$. To integrate $$f(x)$$ over this interval, we can use the power rule of integration:
$\int_{-2}^{1} f(x) \,dx = \int_{-2}^{1} x \,dx = \left[\frac{x^2}{2}\right]_{-2}^{1} = \frac{1^2}{2} - \frac{(-2)^2}{2} = \frac{1}{2} - 2 = - \frac{3}{2}$

2. For $$x \geq 1$$: In this case, the function is $$f(x) = \frac{1}{x}$$. Here, we can integrate using the natural logarithm function:
$\int_{1}^{3} f(x) \,dx = \int_{1}^{3} \frac{1}{x} \,dx = \left[\ln|x|\right]_{1}^{3} = \ln|3| - \ln|1| = \ln(3)$

Now, to find the value of the definite integral over the entire interval $$[-2, 3]$$, we sum up the integrals from both intervals:
$\int_{-2}^{3} f(x) \,dx = \int_{-2}^{1} f(x) \,dx + \int_{1}^{3} f(x) \,dx = -\frac{3}{2} + \ln(3)$

Therefore, the value of the definite integral is $$-\frac{3}{2} + \ln(3)$$.