Question

Find \(\int_0^5 f (x)dx\) if

Short answer

The value of the integral function is \(17.\)

Step-by-step solution

Given information

The integral function is\(\int_0^5 f (x)dx\).

The value of function \(f(x) = 3\) for \(x < 3\) and \(f(x) = x\) for .

Modify the given integral function

Modify the integral function as given below.

\(\int_0^5 f (x)dx = \int_0^3 f (x)dx + \int_3^5 f (x)dx\)

\( = {I_1} + {I_2}\)

Calculate the value of integral function\({I_1}\)using the relation.

\({I_1} = \int_0^3 f (x)dx{\kern 1pt} {\kern 1pt} \,.......(2)\)

So,\(x\)varies from 0 to\(3 .\)

Substitute 3 for\(f(x)\)in Equation (2).

\({I_1} = \int_0^3 3 dx\)

\( = 3(x)_0^3\)

\( = 3(3 - 0)\)

\( = 9{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} .........(3)\)

Calculate the value of integral

Calculate the value of integral function\({I_2}\)using the relation.

\({I_1} = \int_3^5 f (x)dx{\kern 1pt} {\kern 1pt} \,\,{\kern 1pt} .......(4)\)

So,\(x\)varies from 3 to 5.

Substitute\(x\)for\(f(x)\)in Equation (4).

\(\begin{aligned}{l}{I_2} &= \int_3^5 x dx\\ &= \left( {\frac{{{x^2}}}{2}} \right)_3^5\\ &= \frac{1}{2}\left( {{5^2} - {3^2}} \right)\\ &= \frac{1}{2} \times (25 - 9)\end{aligned}\)

\({I_2} = 8{\kern 1pt} \,{\kern 1pt} {\kern 1pt} \,......(5)\)

Calculate the value of the integral function\(\int_0^5 f (x)dx\).

Modify Equation (1) using Equation (3) and Equation (5).

\(\int_0^5 f (x)dx = 9 + 8\)

\( = 17\)

Hence, the value of integral function is 17.