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
I guess your function is
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.
technically, since f(1) is not defined, you need to take the limits as x->1, but you will wind up with the same values.
it saying -8 isn't the number
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 \(-\frac{3}{2} + \ln(3)\).