Problem 10 - Essential Calculus

Question

Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{-2}^{-1}\left(u-\frac{1}{u^{2}}\right) d u $$

Step-by-Step Solution

Answer

The definite integral of the given function from -2 to -1 is \(-1/2\).

Step 1: Split the Integral

Rewrite the integral as the sum of two integrals: \[\int_{-2}^{-1}\left(u-\frac{1}{u^{2}}\right) du = \int_{-2}^{-1} u du - \int_{-2}^{-1}\frac{1}{u^{2}} du\] This simplifies the problem by breaking it into two simpler integrals.

Step 2: Antiderivative of the First Integral

Find the antiderivative of the first integral using the power rule of integration:Given an integral of the form \(\int x^n dx, n \ne -1\), the antiderivative is \(\frac{1}{n+1}x^{n+1}\) + C.So, the antiderivative of \(u\) is \(\frac{1}{2}u^{2}+C_1 \).

Step 3: Antiderivative of the Second Integral

Evaluate the antiderivative of the second integral using the power rule. Given that \( \frac{1}{u^2}=\frac{u^{-2}}{1} \) , let \(n=-2\), then:\(\frac{1}{n+1}u^{n+1} = -u^{-1} + C_2\)The antiderivative of \( \frac{1}{u^2} \) is \(-u^{-1}\) + \(C_2\).

Step 4: Applying Limits to the Antiderivatives

Now substitute the limits into the antiderivatives, applying the fundamental theorem of calculus:The integral from -2 to -1 can be calculated as \(F(-1) - F(-2)\), where \(F(u) = [ \frac{1}{2}u^2 - u^(-1) ]\):\[\int_{-2}^{-1} u du = \frac{1}{2}(-1)^2 - (-1)^{-1} - [\frac{1}{2}(-2)^2 - (-2)^{-1}]\]

Step 5: Calculate the Result

Calculate the result based on above expression:\[= \frac{1}{2} - 1 - [\frac{-1}{2} + 1] = 1 - \frac{3}{2} = -\frac{1}{2}\]