Problem 42 - Applied Calculus

Question

Find the integrals .Check your answers by differentiation. $$\int \frac{e^{\sqrt{y}}}{\sqrt{y}} d y$$

Step-by-Step Solution

Answer

The integral is \(2e^{\sqrt{y}} + C\).

Step 1: Identify the Integral Form

We have the integral \(\int \frac{e^{\sqrt{y}}}{\sqrt{y}} \ d y\). This problem suggests a substitution technique, as the expression includes a form \(e^{\sqrt{y}}\) similar to a function of \(u = \sqrt{y}\).

Step 2: Use Substitution

Set \(u = \sqrt{y}\), which implies \(y = u^2\). Then differentiate \(u = \sqrt{y}\) to get \(dy = 2u \, du\). Substitute these into the integral: \(\int \frac{e^{\sqrt{y}}}{\sqrt{y}} \, dy = \int \frac{e^u}{u} \, 2u \, du = 2 \int e^u \, du\).

Step 3: Integrate with Respect to \(u\)

Integrate \( 2 \int e^u \, du \). This gives \(2 \times e^u\), as the integral of \(e^u\) with respect to \(u\) is \(e^u\). So, the integral becomes \(2e^u + C\), where \(C\) is the constant of integration.

Step 4: Substitute back to \(y\)

Substitute back \(u = \sqrt{y}\) into the integrated result: \(2e^{\sqrt{y}} + C\). This gives the final result of the integration.

Step 5: Verify by Differentiation

Differentiate the result \(2e^{\sqrt{y}} + C\) with respect to \(y\). Using the chain rule, \(\frac{d}{dy} [2e^{\sqrt{y}} + C] = 2 \cdot e^{\sqrt{y}} \cdot \frac{1}{2\sqrt{y}} = \frac{e^{\sqrt{y}}}{\sqrt{y}}\). This matches the original integrand, confirming the solution is correct.

Key Concepts

Integration by Substitution

The substitution method is a powerful tool in calculus, especially when dealing with integrals that are hard to solve directly. With integration by substitution, you make the integral simpler by changing variables. This method is similar to the chain rule in differentiation, where a function is broken into simpler parts.

For example, in the integral \( \int \frac{e^{\sqrt{y}}}{\sqrt{y}} \, dy \), the term \(e^{\sqrt{y}}\) suggests a substitution can make the math more straightforward. You choose a new variable, say \( u \), to represent part of the integrand. Here, letting \( u = \sqrt{y} \) makes sense because it aligns with the exponent and simplifies the expression.

By differentiating \( u = \sqrt{y} \), you'll have \( dy = 2u \, du \), which you then substitute into the integral. Now, the integral becomes \( 2\int e^u \, du \), which is easier to evaluate. This substitution effectively simplifies the integral into a standard form that can be solved directly.

Definite Integrals

Definite integrals are about finding the actual value of the integral over a specified range. The definite integral of a function gives the accumulated sum, taking into account the area under the curve. This is usually expressed as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

In our original problem, the focus is not on a definite integral, but understanding this concept is key when limits are provided. To evaluate it, you first find the indefinite integral, which includes the constant of integration \(C\). Later, for definite integrals, you solve this integral between \(a\) and \(b\), resulting in \(F(b) - F(a)\).

While our exercise simplifies to an indefinite integral, remember that knowing how to handle definite integrals is crucial for applications where specific bounds are involved, like physics and engineering problems.

Chain Rule

The chain rule is a fundamental differentiation technique, used to differentiate compositions of functions. It is instrumental when verifying integrals through differentiation, ensuring the result returns to the original integrand.

In our exercise, after integrating the expression and substituting back to the original variable \(y\), we get \(2e^{\sqrt{y}} + C\). To confirm our solution, we differentiate this expression with respect to \(y\). Here, the chain rule is applied because \(\sqrt{y}\) is a function within another function \(e^{\cdot}\).

Using the chain rule: differentiate \(2e^{\sqrt{y}}\), you obtain \(2 \cdot e^{\sqrt{y}} \cdot \frac{1}{2\sqrt{y}}\). This simplifies back to \(\frac{e^{\sqrt{y}}}{\sqrt{y}}\), the integrand of the original problem. Thus the chain rule helps reaffirm that our integration and substitution process was correct.