# Approximate Integral of sqrt(1 + x^4)

For positive integer values of m and n, functions of the form f(x) = (x^m + 1)^(1/n) don't have elementary antiderivatives unless m and n equal 1 or 2. In particular, the function sqrt(1 + x^4) cannot be integrated exactly using antiderivatives. Instead you must use some approximate methods of integration.

In another tutorial we show how to integrate sqrt(1 + x^4) and sqrt(1 + 9x^4) with series approximations.

In this article we us a different method to estimate definite integrals of sqrt(1 + x^4): bounding it between two functions that do have exact antiderivatives.

## Upper and Lower Bounding Functions for sqrt(x^4 + 1)

Since x^4 is less than x^4 + 1 for all real values of x, we have

x^4 < x^4 + 1

sqrt(x^4) < sqrt(x^4 + 1)

x^2 < sqrt(x^4 + 1)

Thus, the function p(x) = x^2 is a lower bound for the function f(x) = sqrt(x^4 + 1). On the other side, we know that x^4 + 1 + 1/(4x^4) is greater than x^4 + 1 for all real values of x, we also have

x^4 + 1 < x^4 + 1 + 1/(4x^4)

sqrt(x^4 + 1) < sqrt(x^4 + 1 + 1/(4x^4))

sqrt(x^4 + 1) < x^2 + 1/(2x^2)

Thus, the function q(x) = x^2 + 1/(2x^2) is an upper bound for the function f(x) = sqrt(x^4 + 1). Putting it all together we have

x^2 < sqrt(x^4 + 1) < x^2 + 1/(2x^2)

## Finding Bounds on the Integral of sqrt(x^4 + 1)

Since we know that sqrt(x^4 + 1) is bounded between x^2 and x^2 + 1/(2x^2) for all real values of x, definite integrals of sqrt(x^4 + 1) is also bounded between the definite integrals of x^2 and x^2 + 1/(2x^2). Luckily, these bounding functions have elementary antiderivatives, so we can work out the bounds of definite integrals explicitly.

∫ x^2 dx, {x = a to x = b}

= (1/3)x^3, {x = a to x = b}

= (1/3)b^3 - (1/3)a^3

∫ x^2 + 1/(2x^2) dx, {x = a to x = b}

= (1/3)x^3 - 1/(2x), {x = a to x = b}

= (1/3)b^3 - 1/(2b) - (1/3)a^3 + 1/(2a)

In other words, the integral of sqrt(x^4 + 1) on the interval [a,b] is greater than

(1/3)b^3 - (1/3)a^3

but less than

(1/3)b^3 - 1/(2b) - (1/3)a^3 + 1/(2a)

Next we see how to apply this useful bound to estimate integrals of sqrt(x^4 + 1).

## Example 1

Find bounds on the area under the curve y = sqrt(x^4 + 1) from x = 4 to x = 7.

First, using the lower bound x^2 < sqrt(x^4 + 1) we have

∫ x^2 dx, {x = 4 to x = 7}

= (1/3)7^3 - (1/3)4^3

= 279/3

= 93

Next, using the upper bound sqrt(x^4 + 1) < x^2 + 1/(2x^2) we have

∫ x^2 + 1/(2x^2) dx, {x = 4 to x = 7}

= (1/3)7^3 - 1/14 - (1/3)4^3 + 1/8

= 279/3 + 3/56

= 93 ^{3}/_{56}

≈ 93.0536

Therefore, the area under the curve y = sqrt(x^4 + 1) over the interval [4, 7] is between 93 and 93.0536.

## Improving the Lower Bound Function

If the value of |x| is at least 3^(-1/4) ≈ 0.7598, then the lower bound function p(x) = x^2 can be raised to give an even tighter bound. When |x| is at least 3^(-1/4), the following inequality is true

x^4 + 2/3 + 1/(9x^4) ≤ x^4 + 1

Taking the square root of both sides yields

x^2 + 1/(3x^2) ≤ sqrt(x^4 + 1)

Applying this to the definite integral gives us the lower area bound

∫ x^2 + 1/(3x^2) dx, {x = a to x = b}

= (1/3)x^3 - 1/(3x), {x = a to x = b}

= (1/3)[b^3 - a^3 - 1/b + 1/a]

≤ ∫ sqrt(x^4 + 1) dx {x = a to x = b}

## Example 2

Improve the lower bound calculated Example 1.

Since the lower endpoint of the interval x = 4 is larger than 3^(-1/4), we can use the improved lower bound function. Working out the integral gives us

∫ x^2 + 1/(3x^2) dx, {x = 4 to x = 7}

= (1/3)[343 - 64 - 1/7 + 1/4]

= 93 ^{1}/_{28}

≈ 93.0357

Therefore, the area under the curve y = sqrt(x^4 + 1) over the interval [4, 7] is between 93.0357 and 93.0536. Using a numerical integrator, the exact value of the integral of sqrt(x^4 + 1) is approximately 93.0535.

## Extending This Method to Estimate Other Integrals of the Form sqrt(1 + x^m)

Over positive values of x, the function f(x) = sqrt(x^m + 1) is bounded below by p(x) = x^(m/2) and bounded above by q(x) = x^(m/2) + (1/2)x^(-m/2). These bounds are derived from the inequality

x^m < x^m + 1 < x^m + 1 + (1/4)x^-m

Taking the square root of each of the three expressions gives you

x^(m/2) < sqrt(x^m + 1) < x^(m/2) + (1/2)x^(-m/2)

The larger the value of m and the larger the value of x, the tighter the bounds.

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