# How to Solve a Definite Integral

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The solution to a definite integral gives you the area between the integrated function and the x-axis in the Cartesian coordinate system. The lower and upper limits of the interval for the integral represent the left and right bounds of the area. You also use definite integrals in applications such as calculating volume, work, energy and inertia, but first you must learn to apply the basic principles of definite integrals.

• Set up the integral if the problem does not give it to you. If you need to find the area under the curve 3x^2 - 2x + 1 between 1 and 3, for example, you need to take the integral of the function over that interval: int[(3x^2 - 2x + 1)dx] from 1 to 3.

• Use the basic rules of integration to solve the integral the same way you would for an indefinite integral, but do not add the constant of integration. For example, int[(3x^2 - 2x + 1)dx] = x^3 - x^2 + x.

• Substitute the upper limit of the interval of integration for x in the resulting equation and simplify. For example, replacing x with 3 in x^3 - x^2 + x results in 3^3 - 3^2 + 3 = 27 - 9 + 3 = 21.

• Replace x with the lower limit of the interval in the result of the integral and simplify. For example, substituting one into x^3 - x^2 + x gives 1^3 - 1^2 + 1 = 1.

• Subtract the lower limit from the upper limit to get the result of the definite integral. For example, 21 - 1 = 20.

## Tips & Warnings

• To find an area between two curves, subtract the equation for the lower curve from the equation for the upper curve and take the definite integral of the resulting function.
• If the function is discontinuous and the discontinuity lies in the interval of integration, take the definite integral of the first function from the lower limit to the discontinuity and the definite integral of the second function from the discontinuity to the upper limit. Add the results together to get the answer. If the discontinuity is not in the interval of integration, only take the definite integral for the function that exists in the interval.

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