# Derivative of an Integral: Meaning, Formula with Solved Examples

Overview

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The derivative of an integral is what we get when we find the rate of change of the result of an integral. Integration is the process of finding the “anti” derivative, so if we differentiate an integral, we should end up with the original function. However, it’s important to note that this may not always be the case with all definite integrals.

In this mathematics article, we’ll understand the derivative of an integral in various cases, and explore additional examples to enhance our comprehension.

In the case of indefinite integrals, the derivative of an integral function is the function itself. However, it’s important to note that this relationship holds true specifically for indefinite integrals. When dealing with definite integrals, the derivative of an integral function equals the function itself under the condition that the lower limit of integration is a constant and the upper limit is the variable with respect to which we are differentiating. In summary:

Now, we will delve into each of these cases, examining them in detail. Additionally, we will explore the evaluation of definite integrals that do not fall into either of the preceding two cases.

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The derivative of an indefinite integral represents the original function from which the integral was calculated. In other words, it allows us to “undo” the integration process and find the original function.

To illustrate this concept, let’s consider an example. Suppose we have the following indefinite integral:

$$\int(3x^{2} + 2x)dx$$

To find the derivative of this integral, we differentiate each term separately using the power rule of differentiation. The power rule states that for a term of the form $$x^{n}$$, the derivative is given by $$nx^{(n-1)}$$.

Differentiating the terms in our integral, we get:

$$\frac{d}{dx}(3x^{2}) = 6x$$

$$\frac{d}{dx}(2x) = 2$$

Now, combining the derivatives of each term, we obtain the derivative of the original integral:

$$\frac{d}{dx} \int (3x^{2} + 2x).dx = 6x + 2$$

So, the derivative of the indefinite integral $$\int(3x^{2} + 2x)dx$$ is $$6x + 2$$. This means that if we integrate $$6x + 2$$, we will retrieve the original function $$3x^{2} + 2x$$.

The derivative of a definite integral refers to finding the rate of change of the integral with respect to a variable. In simpler terms, it involves determining how the value of the integral changes as the variable in the limits of integration varies.

The format of a definite integral is $$\int_{a}^{b}f(t)dt$$. However, it’s important to note that the limits of integration don’t always have to be constants. In fact, there are three possible cases:

In this article, we will explore each of these cases and learn how to find the derivative of an integral in each scenario.

Let’s consider the definite integral $$\int_{a}^{b}f(x)dx$$, where ‘$$a$$’ and ‘$$b$$’ are constants. According to the second fundamental theorem of calculus, we have $$\int_{a}^{b}f(x)dx = F(b) – F(a)$$, where $$F(x)$$ is the antiderivative of $$f(x)$$, given by $$\int f(t)dt$$.

Now, let’s find the derivative of this integral. Taking the derivative with respect to $$x$$, we get $$\frac{d}{dx}\int_{a}^{b}f(x)dx = \frac{d}{dx}[F(b) – F(a)] = 0$$ (since $$F(b)$$ and $$F(a)$$ are constants). Therefore, when both limits are constants, the derivative of a definite integral is $$0$$, i.e., \frac{d}{dx}\int_{a}^{b}f(t)dt = 0.

Let’s consider a definite integral $$\int_{a}^{x}f(t).dt$$, where ‘$$a$$’ is a constant and ‘$$x$$’ is a variable. According to the first fundamental theorem of calculus, the derivative of this integral, $$\frac{\mathrm{d}}{\mathrm{d}x}\int_{a}^{x}f(t).dt$$, is equal to the original function, $$f(x)$$. This fundamental principle highlights the fact that the derivative of an integral is the original function itself. Here are a few examples to illustrate this concept:

It’s important to note that when the lower limit of integration is a constant and the upper limit is a variable, we can rewrite the integral accordingly. For instance, consider $$\frac{d}{dx}\int_{2}^{x}t^{3}dt$$. To handle this situation, we can utilize the property of the definite integral and rewrite it as $$-\frac{d}{dx}\int_{2}^{x}t^{3}dt$$. Applying the examples mentioned earlier, the result would be $$-x^{3}$$.

These examples demonstrate the fundamental principle that the derivative of an integral yields the original function. It’s crucial to pay attention to the limits of integration and correctly handle situations where the lower or upper limit varies.

When dealing with a definite integral where both limits of integration involve variables, finding the derivative can be a bit trickier. Let’s explore this concept with a simple example to better understand it.

Consider the integral: $$I = \int_{a(x)}^{b(x)}f(t).dt$$, where both ‘$$a(x)$$’ and ‘$$b(x)$$’ are functions of ‘$$x$$’ instead of constant values. We want to find the derivative of this integral with respect to ‘$$x$$’, denoted as $$\frac{dI}{dx}$$.

To solve this, we’ll employ the Leibniz rule, which states that for a function with varying limits of integration, the derivative can be obtained by differentiating both the integrand and the limits of integration.

Let’s work through an example:

Example: Find the derivative of the integral $$I = \int_{x}^{x^{2}}t^{3}.dt$$.

Solution: To find $$\frac{dI}{dx}$$, we differentiate both the integrand and the limits of integration with respect to ‘$$x$$’.

Therefore, the derivative of the given integral is $$\frac{dI}{dx} = 6x^{5} – 3x^{3}$$.

The formula of derivative of the integral is given as follows:

$$\int_{g(t)}^{h(t)}f(x)dx = h'(t)\times f(h(t)) – g'(t)\times f(g(t))$$

where, $$f(h(t))$$ and $$f(g(t))$$ are the composite functions.

To find the derivative of an integral using the above formula, follow the below steps:

Step 1: Start by calculating the derivative of the upper limit and substitute it into the integrand. Multiply both results.

Step 2: Next, find the derivative of the lower limit and substitute it into the integrand. Multiply both results.

Step 3: Finally, subtract the results obtained in the previous two steps.

By following these steps, you can effectively compute the derivative of an integral.

The derivative of an integral finds numerous applications across various fields. Here are some key applications:

1.Physics:

2.Economics and Finance:

3.Engineering:

4.Probability and Statistics:

1) Find the derivative of the function $$f(x) = \int_{0}^{x} e^{t}dt$$.

Solution)

To find the derivative of this integral function, we can apply the fundamental theorem of calculus.

According to the theorem, if we have an integral function $$F(x) = \int_{a}^{x}f(t)dt$$, then its derivative is given by $$F'(x) = f(x)$$.

So, the integral function is $$F(x) = \int_{0}^{x} e^{t}dt$$.

By applying the theorem, we find that the derivative of $$F(x)$$ is $$f(x) = e^{x}$$.

Therefore, the derivative of $$f(x) = \int_{0}^{x} e^{t}dt$$ is $$f'(x) = e^{x}$$.

2) Find the derivative of the function $$f(x) = \int (2x^{3} – 4x + 1) dx$$.

Solution)

To find the derivative of an indefinite integral, we can use the fundamental theorem of calculus.

According to the theorem, if we have an indefinite integral function $$F(x) = \int f(x).dx$$, then its derivative is given by $$F'(x) = f(x)$$.

So, the indefinite integral function is $$F(x) = \int (2x^{3} – 4x + 1) dx$$.

By applying the theorem, we find that the derivative of $$F(x)$$ is $$f(x) = 2x^{3} – 4x + 1$$.

Therefore, the derivative of $$f(x) = \int (2x^{3} – 4x + 1)dx$$ is $$f'(x) = 2x^{3} – 4x + 1$$.

We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.