If f(x) satisfies f(1) = f(4), then what is $\rm \int^4_1f'(x) dx$ equal to?

Question

If f(x) satisfies f(1) = f(4), then what is $\rm \int^4_1f'(x) dx$  equal to?

Prepp Team · Accepted Answer

The question asks us to evaluate the definite integral of the derivative of a function, given a specific condition about the function's values at the integration limits.

Understanding the Integral and the Fundamental Theorem of Calculus

The expression $\rm \int^4_1f'(x) dx$ represents the definite integral of the function $f'(x)$ with respect to $x$, from the lower limit 1 to the upper limit 4.

This type of integral is directly related to the original function $f(x)$ through the Fundamental Theorem of Calculus, Part 2. This theorem provides a way to evaluate definite integrals if we know an antiderivative of the function being integrated.

The Fundamental Theorem of Calculus (Part 2) states that if $F(x)$ is an antiderivative of a continuous function $g(x)$ on the interval $[a, b]$, then:

$\rm \int_a^b g(x) dx = F(b) - F(a)$

In our case, the function being integrated is $f'(x)$. An antiderivative of $f'(x)$ is the original function $f(x)$, because the derivative of $f(x)$ is $f'(x)$. Here, $g(x) = f'(x)$ and $F(x) = f(x)$. The limits of integration are $a=1$ and $b=4$.

Applying the Theorem to the Given Integral

Using the Fundamental Theorem of Calculus, Part 2, we can evaluate the integral $\rm \int^4_1f'(x) dx$ as the difference of the antiderivative $f(x)$ evaluated at the upper and lower limits:

$\rm \int^4_1f'(x) dx = [f(x)]^4_1 = f(4) - f(1)$

So, the value of the integral is equal to the value of the function $f(x)$ at $x=4$ minus the value of the function $f(x)$ at $x=1$.

Using the Condition f(1)=f(4)

The problem provides the condition that $f(1) = f(4)$. This means the value of the function at the lower limit of integration is the same as its value at the upper limit of integration.

We found that the integral is equal to $f(4) - f(1)$. Let's substitute the given condition $f(1) = f(4)$ into this expression:

$\rm \int^4_1f'(x) dx = f(4) - f(1)$

Since $f(1) = f(4)$, we can replace $f(1)$ with $f(4)$ (or $f(4)$ with $f(1)$):

$\rm f(4) - f(1) = f(4) - f(4)$

Subtracting a value from itself results in zero:

$\rm f(4) - f(4) = 0$

Therefore, the value of the integral $\rm \int^4_1f'(x) dx$ is 0.

Conclusion and Final Answer

By applying the Fundamental Theorem of Calculus, we found that the definite integral $\rm \int^4_1f'(x) dx$ is equal to $f(4) - f(1)$. Given the condition $f(1) = f(4)$, this difference is $f(4) - f(4) = 0$.

The value of $\rm \int^4_1f'(x) dx$ is 0.

Revision Table: Key Calculus Concepts

Concept	Description	Formula/Notation
Derivative	Measures the instantaneous rate of change of a function.	$f'(x)$ or $\frac{dy}{dx}$
Antiderivative	A function $F(x)$ whose derivative is the given function $f(x)$.	If $F'(x) = f(x)$, then $F(x)$ is an antiderivative of $f(x)$.
Definite Integral	Represents the net signed area under the curve of a function between two limits, or the total change of a quantity whose rate of change is given by the function.	$\rm \int_a^b f(x) dx$
Fundamental Theorem of Calculus (Part 2)	Relates definite integrals to antiderivatives, providing a method for evaluating definite integrals.	$\rm \int_a^b f(x) dx = F(b) - F(a)$, where $F'(x)=f(x)$.

Additional Information: Properties of Definite Integrals

Understanding definite integrals involves knowing their properties. Some key properties include:

Linearity: $\rm \int_a^b [cf(x) + dg(x)] dx = c\int_a^b f(x) dx + d\int_a^b g(x) dx$ (where c and d are constants)
Additivity over the Interval: $\rm \int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x) dx$ (for $a < c < b$)
Reversing Limits: $\rm \int_a^b f(x) dx = -\int_b^a f(x) dx$
Integral with Same Limits: $\rm \int_a^a f(x) dx = 0$

These properties are useful when manipulating or evaluating definite integrals in various calculus problems.

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