Derivative of an Integral

The derivative of an integral is the result obtained by differentiating the result of an integral. Integration is the process of finding the "anti" derivative and hence by differentiating an integral should result in the original function itself. But this may not be the scenario with all definite integrals.

Let us learn more about the derivative of an integral (in different cases) along with more examples.

The derivative of an integral of a function is the function itself. But this is always true only in the case of indefinite integrals. The derivative of a definite integral of a function is the function itself only when the lower limit of the integral is a constant and the upper limit is the variable with respect to which we are differentiating. To summarize:

Let us look into each of these cases in detail. Also, let us see how to evaluate definite integrals that don't match with any of the last two cases.

Let us consider an indefinite integral ∫ x3/2 dx. If we evaluate this using the power rule of integration, we get (2x5/2)/5 + C. If we differentiate this using the power rule of differentiation, we get (2/5) (5/2) (x3/2) + 0 = x3/2. If we put this integration and differentiation in one step, we can write it as

d/dx ∫ x3/2 dx = x3/2.

i.e., the derivative of an indefinite integral is equal to the original function itself (it is a kind of "derivative and integral symbol get canceled with each other"). Thus, for any function f(x), we can write

d/dx ∫ f(x) dx = f(x)

A definite integral is of the form ∫a b f(t) dt. But here, the limits always don't need to be constants. There can be 3 cases.

We will study how to find the derivative of an integral in each of these cases.

Consider the definite integral ∫a b f(x) dx where both 'a' and 'b' are constants. Then by the second fundamental theorem of calculus, ∫a b f(x) dx = F(b) - F(a) where F(x) = ∫ f(t) dt. Now, let us compute its derivative. d/dx∫a b f(x) dx = d/dx [F(b) - F(a)] = 0 (as F(b) and F(a) are constants). Thus, when both limits are constants, the derivative of a definite integral is 0.

i.e., d/dx ∫a b f(t) dt = 0

Consider a definite integral ∫ax f(t) dt, where 'a' is a constant and 'x' is a variable. Then by the first fundamental theorem of calculus, d/dx ∫ax f(t) dt = f(x). This would reflect the fact that the derivative of an integral is the original function itself. Here are some examples.

Note that here the lower limit should be a constant. If the upper limit is a constant and the lower limit is a variable, say d/dx ∫x2 t3 dt then we rewrite (using the property of definite integral) as - d/dx ∫2x t3 dt and now its result using the above examples is -x3.

Consider the integral ∫t²t³ log (x3 + 1) dx. Here, both the limits involve the variable t. In such cases, we apply a property of definite integral that says ∫ac f(t) dt = ∫ab f(t) dt + ∫bc f(t) dt and we assume 'b' to be a random constant while applying this property. Then we can write the above integral as

∫t²t³ log (x3 + 1) dx = ∫t²1 log (x3 + 1) dx + ∫1t³ log (x3 + 1) dx

Now, we apply another property of definite integrals that says ∫ab f(t) dt = - ∫ba f(t) dt. Using this, we can write ∫t²1 log (x3 + 1) dx as - ∫1t² log (x3 + 1) dx. Now the above step becomes:

= - ∫1t² log (x3 + 1) dx + ∫1t³ log (x3 + 1) dx

Taking the derivative on both sides,

d/dt ∫t²t³ log (x3 + 1) dx = - d/dt ∫1t² log (x3 + 1) dx + d/dt ∫1t³ log (x3 + 1) dx

For the first integral, assume that t2 = u and for the second integral, assume that t3 = v. By the chain rule, we can write

= - [ d/du ∫1u log (x3 + 1) dx ] [ du/dt ] + [ d/dv ∫1v log (x3 + 1) dx ] [ dv/dt ]

= - [ d/du ∫1u log (x3 + 1) dx ] [ 2t ] + [ d/dv ∫1v log (x3 + 1) dx ] [ 3t2 ]

Since the each integral has its lower limit to be a constant and upper limit to be a variable, their derivatives are equal to the functions in terms of the respective variables. i.e.,

= [- log (u3 + 1) ] [ 2t ] + [log (v3 + 1) ] [ 3t2 ]

Substitute u = t2 and v = t3 back here,

= - 2t log (t6 + 1) + 3t2 log (t9 + 1)

This procedure is very helpful to find out the derivative of an integral.

We have seen that the derivative of the integral ∫t²t³ log (x3 + 1) dx is - 2t log (t6 + 1) + 3t2 log (t9 + 1) and this can be written as 3t2 log (t9 + 1) - 2t log (t6 + 1). Note that the derivative of the upper limit t3 is 3t2 and the derivative of the lower limit t2 is 2t here. Thus, we can compute the derivative of an integral formula as follows:

∫g(t)h(t) f(x) dx = h'(t) · f(h(t)) - g'(t) · f(g(t))

where, f(h(t)) and f(g(t)) are the composite functions. i.e., to find the derivative of an integral:

☛Related Topics:

Example 1: Find the derivative of the integral ∫-12 sin x dx without actually evaluating the integral.

Solution:

We know that the definite integral with constant limits leads to a constant.

Also, we know that the derivative of the constant is 0.

Hence, d/dx ∫-12 sin x dx = 0

Answer: 0.

Example 2: Prove the result of Example 1 by evaluating the integral.

Solution:

Let us compute the given integral.

∫-12 sin x dx = (- cos x)-12

= - cos 2 + cos (-1)

Now, we will compute the derivative of the given integral.

d/dx ∫-12 sin x dx = d/dx [- cos 2 + cos (-1)]

= 0 + 0

= 0

Answer: The result of Example 1 is verified.

Example 3: What is the derivative of the integral ∫x2x sin √t dt?

Solution:

The given integral can be written as:

∫x2x sin √t dt = ∫x0 sin √t dt + ∫02x sin √t dt

= - ∫0x sin √t dt + ∫02x sin √t dt

We will take the derivative of integrals.

d/dx ∫x2x sin √t dt = - d/dx ∫0x sin √t dt + d/dx ∫02x sin √t dt

For the second integral, assume that 2x = u. The second integral becomes d/dx ∫0u sin √t dt and by chain rule it can be written as (d/du ∫0u sin √t dt) · (du/dx) = (d/du ∫0u sin √t dt) (2). Then the above step can be rewritten as:

d/dx ∫x2x sin √t dt = - d/dx ∫0x sin √t dt + 2 d/du ∫0u sin √t dt

Now, using the lower limit of each of these integrals is a constant. So as explained earlier, its derivative is the function (in terms of the upper limit) itself. So we get

d/dx ∫x2x sin √t dt = - sin √x + 2 sin √u

Substitute u = 2x back,

d/dx ∫x2x sin √t dt = - sin √x + 2 sin √2x

Answer: - sin √x + 2 sin √2x

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The derivative of an integral is the function itself when the lower limit of the integral is a constant and the upper limit is just a variable. i.e., d/dx ∫ax f(t) dt = f(x), where 'a' is a constant.

To find the derivative of an integral when both the limits of a definite integral are not constants, then we apply the following two properties to split the given integral into two integrals where each of them has its lower limit to be a constant.

∫ac f(t) dt = ∫ab f(t) dt + ∫bc f(t) dt

∫ab f(t) dt = - ∫ba f(t) dt

Also, we apply the substitution method of integration if the upper limit of each integral is not just a variable. Then the derivative of each of the integrals is the function itself in terms of its respective upper bound.

For differentiating integrals:

If the integral has no limits, then its derivative is the actual function itself. For example, d/dx ∫ f(x) dx = f(x). This is because integration is just the reverse process of differentiation.

Yes, the integral of a derivative is the function itself, but an added constant may vary. For example, d/dx (x2) = 2x, where as ∫ d/dx (x2) dx = ∫ 2x dx = 2(x2/2) + C = x2 + C. Here the original function was x2 whereas the integral is x2 + C, where C is the integration constant.

No, the derivative of an integral of a function doesn't need to equal to the function itself. It happens only when the integral is an indefinite integral or it is a definite integral with lower bound to be a constant and upper bound to be a variable.