Integration of the Square Root Function

By Karl Wallulis
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Integration of the square root function contains the same procedures as integrating any polynomial equation. The square root operation is the same as the 1/2 power, so you can use the power rule for integrals to find the antiderivative. If necessary, use u-substitution to rewrite the radical expression before integrating.

The Power Rule

The square root function is defined as f(x) = A*Sqrt(Bx + C) + D, where A, B, C and D are all real numbers. Since the square root is equivalent to a power of 1/2, the function can be integrated using the power rule for integrals. The power rule states that any function f(x) = x^(n) has an antiderivative F(x) = (x^(n + 1)) / (n + 1). The antiderivative of f(x) = Sqrt(x) is, therefore, F(x) = (x^(1/2 + 1)) / (1/2 + 1), which simplifies to 2x^(3/2) / 3.


The power rule for square roots only works when the expression under the radical symbol is simply "x." For other square root functions, such as f(x) = Sqrt(2x + 1), it is necessary to perform u-substitution, replacing 2x + 1 with a variable u, then rewriting the rest of the integral in terms of u. This includes changing the dummy variable dx to du and changing the bounds from x = a and x = b to u = c and u = d.

For example, after replacing 2x + 1 with u in the integral from 1 to 8 of f(x) = Sqrt(2x + 1)dx, rewrite dx as du by differentiating the equation u = 2x + 1 to get du = 2 dx, or dx = 1/2 du. Then rewrite the bounds by solving for the value of u when x = 1 and x = 8: u = 2(1) + 1 or u = 3 and u = 2(8) + 1 or u = 17 are the new bounds. The integral is now from 3 to 17 of f(u) = 1/2 * Sqrt(u) du.

Indefinite Integral

There are two types of integrals: definite integrals, which have bounds delimiting the area of integration, and indefinite integrals, which have no bounds. To calculate the indefinite integral of a square root function, use u-substitution if necessary, calculate the antiderivative of the function, substitute the value of x back in for u and add the constant of integration C. For the function f(x) = Sqrt(2x + 1) dx, the antiderivative of 1/2 Sqrt(u) is equal to 1/2* 2u^(2/3) / 3 or u^(2/3) / 3. Substituting "2x + 1" back into the equation and adding the constant of integration yields the antiderivative F(x) = ((2x + 1)^2/3) / 3 + C.

Definite Integral

Unlike indefinite integrals, definite integrals have bounds; therefore, the solution is not an equation but rather a value that corresponds to the area under the curve between the two bounds. Calculate the value of a definite integral by evaluating the antiderivative at the upper and lower bound of the integral and subtracting the value at the lower bound from the value at the upper bound. For example, if the integral of f(x) = Sqrt(2x + 1) dx is from x = 0 to x = 2, evaluate F(2) - F(0). F(2) = (2(2) + 1)^(2/3) / 3 + C and F(1) = (2(0) + 1)^(2/3) / 3 + C, so the value of the integral is approximately 0.97 - 0.67 or 0.3.

About the Author

Karl Wallulis has been writing since 2010. He has written for the Guide to Online Schools website, covering academic and professional topics for young adults looking at higher-education opportunities. Wallulis holds a Bachelor of Arts in psychology from Whitman College.