Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{-1}^{1} \frac{1}{x+2} d x $$

Short answer

The value of the definite integral \(\int_{-1}^{1} \frac{1}{x+2} dx\) is \(\ln3\).

Step-by-step solution

Calculate the antiderivative

The first step is to find the antiderivative of the function \(f(x)=\frac{1}{x+2}\). Using the rule of logarithmic integration, we know that the antiderivative of \(f(x)\) is \(\ln|x+2|\) plus a constant \(C\). So our indefinite integral is \(\int \frac{1}{x+2} d x = \ln|x+2| + C\).

Apply the Fundamental theorem of Calculus

The fundamental theorem of calculus states that the definite integral of a function over an interval from \(a\) to \(b\) is equal to the antiderivative evaluated at \(b\) minus the antiderivative evaluated at \(a\). In this instance, the interval is from \(-1\) to \(1\). Substituting these values into the antiderivative, \(\ln|x+2|\), calculated in step 1, we get \(\ln|1+2| - \ln|-1+2| = \ln|3| - \ln|1|\).

Simplify the result

Simplify the results from step 2 yields \(\ln3 - \ln1 = \ln3 - 0 = \ln3\). Alanother way to think about this is that the quantity \(\ln|-1+2|\) equals \(\ln1\) which is equal to zero given that any number to the power zero equals one.

Verify using a graphing utility

The exercise asks to use a graphing utility to verify the result. Pull up a graphing tool and input the function \(f(x)=\frac{1}{x+2}\) and evaluate the area under the curve from \(-1\) to \(1\). The area should match with the result obtained \(\ln3\) demonstrating that the manual calculation is correct.