6.2
Definite integrals of products of two functions over fixed intervals can be solved using integration by parts.
To solve the right-hand side of this expression, the difference of the product of functions will be evaluated between the interval’s endpoints, and the remaining term will be treated as a definite integral.
A useful example is the integral of the inverse tangent function. Because no standard integral formula exists for this function, the integrand is instead treated as a product of the inverse tangent and the constant 1.
The inverse tangent is taken as the function to be differentiated, and the constant is integrated.
Substituting into the integration by parts formula, the first term can be solved by directly evaluating the product at the endpoints. The remaining integral can then be solved by substitution.
A new variable, t, is set equal to 1 plus x squared and the limits of integration are adjusted. The integral then becomes a reciprocal expression that simplifies to a logarithmic form.
The natural log of one equals zero, so the final expression simplifies to show the area under the curve between the two limits.
Definite integrals involving the product of two functions over a fixed interval can be evaluated using integration by parts. This method rewrites the…
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