Question

the function f is defined by f(x) = {2 for x<3, x-1 for x>3. what is the value of the integral from 1 to 5 of f(x) dx? I know the answer is 10 but I don't know how?

Answer 1

Since

$f(x)=\begin{cases}2&\text{for }x<3\\x-1&\text{for }x\ge3\end{cases}$

you have

$\displaystyle\int_1^5f(x)\,\mathrm dx=\int_1^32\,\mathrm dx+\int_3^5(x-1)\,\mathrm dx$

You have to split up the integral like this because $f(x)$ is not the same over the subintervals [1, 3) and [3, 5]. Proceed with integration as usual:

$\displaystyle\int_1^32\,\mathrm dx=2x\bigg|_{x=1}^{x=3}=2(3-1)=4$
$\displaystyle\int_3^5(x-1)\,\mathrm dx=\frac12x^2-x\bigg|_{x=3}^{x=5}=\left(\frac{5^2}2-5\right)-\left(\frac{3^2}2-3\right)=6$

So $\displaystyle\int_1^5f(x)\,\mathrm dx=4+6=10$

Answer 2

The value of the given function after integrating under the limits from 1 to 5 is 10 and this can be determined by using the given data.

Given :

The function f is defined by f(x) = {2 for x < 3, x-1 for x > 3.

Integrate the given function:

$\rm \int\limits^5_1 {f(x)} \, dx =\int\limits^3_1 {2} \, dx + \int\limits^5_3 {(x-1)} \, dx$

$\rm \int\limits^5_1 {f(x)} \, dx =\left[2x\right]^3_1 +\left[\dfrac{x^2}{2}-x\right]^5_3$

Simplify the above expression by putting the limits.

$\rm \int\limits^5_1 {f(x)} \, dx =\left[2(3)-2(1)\right] +\left[\dfrac{5^2}{2}-5-\dfrac{3^2}{2}+3\right]$

Now, apply arithmetic operations to solve the above expression.

$\rm \int\limits^5_1 {f(x)} \, dx =\left[6-2\right] +\left[\dfrac{25}{2}-5-\dfrac{9}{2}+3\right]$

$\rm \int\limits^5_1 {f(x)} \, dx =\left[6-2\right] +\left[\dfrac{25}{2}-5-\dfrac{9}{2}+3\right]$

$\rm \int\limits^5_1 {f(x)} \, dx =4+6 = 10$

The value of the given function after integrating under the limits from 1 to 5 is 10 and this can be determined by using the given data.

For more information, refer to the link given below:

https://brainly.com/question/14502499