Question

If \(F(x) = \int_2^x f (t)dt\), where \(f\) is the function whose graph is given, which of the following values is largest?

(A) \(F(0)\)

(B) \(F(1)\)

(C) \(F(2)\)

(D) \(F(3)\)

(E) \(F(4)\)

Short answer

The largest integral function is \(F(2)\).

Step-by-step solution

Property Used for the integral

Show the Property of definite integral

\(\int_a^b f (x)dx = - \int_b^a f (x)dx{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,{\kern 1pt} .......(1)\)

Step 2: Calculate the value of the integral function at \(x = 0\)

The graph of the function\(f\).

The integral function is\(F(x) = \int_2^x f (t)dt{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ......(2)\)

Calculate the value of the integral function at\(x = 0\)using Equation (2).

Substitute 0 for\(x\)in Equation (1)

\(F(0) = \int_2^0 f (t)dt{\kern 1pt} \,{\kern 1pt} {\kern 1pt} ......(3)\)

Apply the Property of Definite integral.

Modify Equation (3) using Equation (1).

\(F(0) = - \int_0^2 f (t)dt\)

Hence, the value of \(F(0)\) is negative.

Calculate the value of the integral function at \(x = 1\)

Calculate the value of the integral function at\(x = 1\)using Equation (2)

Substitute 1 for\(x\)in Equation (1).

\(F(1) = \int_2^1 f (t)dt{\kern 1pt} {\kern 1pt} {\kern 1pt} \,.......(4)\)

Apply the Property of the Definite integral.

Modify Equation (4) using Equation (2).

\(F(1) = - \int_1^2 f (t)dt\)

Hence, the value of \(F\left( 1 \right)\) is negative.

Calculate the value of the integral function at \(x = 2\)

Calculate the value of the integral function at\(x = 2\)using Equation (2).

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

\(F(2) = \int_2^2 f (t)dt{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \,........(5)\)

The value of\(F(2)\)is a integral with width of the interval\(\Delta x = 0\).

So, the value of \(F(2)\) is \(0.\)

Calculate the value of the integral function at \(x = 3\)

Calculate the value of the integral function at\(x = 3\)using Equation (2).

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

\(F(1) = \int_2^3 f (t)dt\)

The area of the function within limits\((2,3)\)is negative as it is below the\(x\)-axis.

So, the value of \(F(3) = \int_2^3 f (t)dt\) and is negative.

Calculate the value of the integral function at \(x = 4\)

Calculate the value of the integral function at\(x = 4\)using Equation (2).

Substitute 4 for\(x\)in Equation (2)

\(F(4) = \int_2^4 f (t)dt\)

The area of the function within limits\((2,4)\)is negative as it is below the\(x\)-axis.

So, the value of \(F(4)\) is negative.

Comparison of the values

The value of\(F(0),F(1),F(2),F(3)\)and\(F(4)\). The values of\(F(0),F(1),F(3)\), and\(F(4)\)is negative. The value of\(F(2)\)is zero and non-negative.

Hence, the value of \(F(2)\) is the largest among the given functions.