Question

Show that for any number \(\mathrm{a}>1\) \(\int_{1}^{a} \ln x d x+\int_{0}^{\ln a} \mathrm{e}^{y} d y=a \ln a\)

Short answer

Question: Show that for any number \(a > 1\), \(\int_{1}^{a} \ln x dx + \int_{0}^{\ln a} e^y dy = a \ln a\).

Step-by-step solution

Evaluate the first integral \(\int_{1}^{a} \ln x dx\)

To evaluate this integral, we'll use integration by parts. Integration by parts states that \(\int u dv = uv - \int v du\). We'll choose \(u = \ln x\) and \(dv = dx\). So, \(du = \frac{1}{x}dx\) and \(v = x\). Now, substituting these into the integration by parts formula, we get: \(\int \ln x dx = x \ln x - \int x \cdot \frac{1}{x} dx\) Now we can evaluate this integral: \(\int \ln x dx = x \ln x - \int dx = x \ln x - x + C\) We can now evaluate the definite integral: \(\int_{1}^{a} \ln x dx = \left[a \ln a - a\right] - \left[\ln 1 - 1\right] = a\ln a - a + 1\)

Evaluate the second integral \(\int_{0}^{\ln a} e^y dy\)

To evaluate the integral \(\int e^y dy\), we know that the integral of \(e^y\) with respect to \(y\) is just \(e^y\): \(\int e^y dy = e^y + C\) Now, we can evaluate the definite integral: \(\int_{0}^{\ln a} e^y dy = \left[e^{\ln a}\right] - \left[e^{0}\right] = a - 1\)

Sum the two integrals and check if it equals \(a \ln a\)

Now, let's sum the two integrals and check if it equals \(a \ln a\): \((a\ln a - a + 1) + (a - 1) = a\ln a - a + 1 + a - 1 = a\ln a\) Thus, we have shown that for any number \(a > 1\), \(\int_{1}^{a} \ln x dx + \int_{0}^{\ln a} e^y dy = a \ln a\).