Question

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

Short answer

Question: Show that the integral of ln(x) on the interval [0,1] using integration by parts is equivalent in magnitude but opposite in sign to the integral \(-\int_{0}^{\infty} e^{-x} \, dx\). Solution: Based on our calculation using integration by parts, we found that $$\int_{0}^{1} \ln(x) \, dx = -1$$ and $$-\int_{0}^{\infty} e^{-x} \, dx = 1$$ These two integral results are equal in magnitude but have opposite signs.

Step-by-step solution

Choose u(x) and v'(x)

To use integration by parts, we first pick u(x) and v'(x). We want to choose these functions such that the integral of u'(x)v(x) is easier to compute than the original integral. It's common to choose u(x) = ln(x) and v'(x) = 1.

Compute u'(x) and v(x)

Next, we need to compute the derivatives and antiderivatives for our chosen functions u(x) = ln(x) and v'(x) = 1. For u(x) = ln(x), the derivative u'(x) is: $$u'(x) = \frac{1}{x}$$ For v'(x) = 1, the antiderivative v(x) is: $$v(x) = x$$

Apply the integration by parts formula and evaluate on the interval [0,1]

Now, we will apply the integration by parts formula using our chosen functions and their derivatives: $$\int_{0}^{1} \ln(x) \, dx = \int_{0}^{1} u(x) v'(x) \, dx = \bigg[u(x)v(x) - \int_{0}^{1} u'(x)v(x) \, dx\bigg] $$ Substitute u(x), u'(x), and v(x) from Step 2: $$= \bigg[\ln(x)x - \int_{0}^{1} \frac{1}{x}x \, dx\bigg]$$ Now, evaluate the integral on the interval [0,1]: $$\bigg[\ln(1)(1) - \ln(0)(0) - \int_{0}^{1} 1 \, dx\bigg] = -\int_{0}^{1} 1 \, dx$$ The integral of 1 is x, so we get: $$-\Bigg[x\Bigg]_0^1 = -(1 - 0) = -1\text.$$ Thus, $$\int_{0}^{1} \ln(x) \, dx = -1\text.$$

Compute the second integral and compare results

Now we need to compute the given integral \(-\int_{0}^{\infty} e^{-x} \, dx\). We find the antiderivative of \(e^{-x}\): $$\int e^{-x} \, dx = -e^{-x}$$ Evaluate the integral on the interval [0,∞]: $$-\Bigg[-e^{-x}\Bigg]_0^\infty = -\Big(-e^{-\infty} + e^0\Big) = 1-0 = 1$$ Now we have the results for both integrals: $$\int_{0}^{1} \ln(x) \, dx = -1$$ $$-\int_{0}^{\infty} e^{-x} \, dx = 1$$ As the results are negative of each other, we can conclude that they are equivalent in magnitude, but opposite in sign.