Chapter 7: Problem 38
Find the integrals Check your answers by differentiation. $$\int \frac{e^{\sqrt{y}}}{\sqrt{y}} d y$$
Short Answer
Expert verified
The integral is \(2e^{\sqrt{y}} + C\). Verification by differentiation confirms this is correct.
Step by step solution
01
Identify substitution
We recognize that the integrand involves a composition with a square root function. We can simplify the integral by using substitution. Let \( u = \sqrt{y} \). Then, \( y = u^2 \) and \( dy = 2u \, du \). This substitution will simplify the given integral.
02
Rewrite the integral using substitution
Substituting \( u = \sqrt{y} \) into the integral, we have: \[ \int \frac{e^{u}}{u} \cdot 2u \, du \].This simplifies to: \[ 2 \int e^u \, du \].
03
Integrate the simplified expression
Now, we integrate the simpler expression \( 2 \int e^u \, du \). The antiderivative of \( e^u \) is \( e^u \). Therefore, \[ 2 \int e^u \, du = 2e^u + C \].
04
Substitute back to the original variable
Substitute \( u = \sqrt{y} \) back into the result of the integration. The final result is: \[ 2e^{\sqrt{y}} + C \]. This is the integral of the original function.
05
Differentiate the result to verify
To check our answer, we differentiate the result: \( 2e^{\sqrt{y}} + C \). The derivative gives us:\[ \frac{d}{dy} [2e^{\sqrt{y}}] = 2 \cdot e^{\sqrt{y}} \cdot \frac{1}{2\sqrt{y}} = \frac{e^{\sqrt{y}}}{\sqrt{y}} \].This matches the original integrand, confirming that the integration was done correctly.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a popular technique in calculus used to simplify complex integrals. The technique involves replacing a complicated section of an integral with a single variable, making it easier to evaluate.
Here's how it generally works:
Here's how it generally works:
- Identify a portion of the integrand that, when differentiated, appears elsewhere in the integral. This is usually called the "inner function."
- Choose a new variable, let's say \( u \), to represent this part of the integrand (for example, \( u = \sqrt{y} \)).
- Express the original variable (in this case, \( y \)) in terms of \( u \) (so \( y = u^2 \)).
- Calculate the differential \( dy \) in terms of \( du \) (from \( y = u^2 \), \( dy = 2u \, du \)).
Exponential Functions
Exponential functions are a crucial part of calculus and often appear in integrals and derivatives. A basic exponential function has the form \( f(x) = a e^{bx} \), where \( e \) is the base of the natural logarithm, and it is approximately equal to 2.718.
- These functions grow rapidly and are unique due to their specific property where their rate of growth is proportional to their value.
- This means that derivatives and integrals involving exponentials often maintain the same form.
- For example, the derivative and the integral of \( e^x \) are both \( e^x \), making it relatively simple to work with.
Differentiation
Differentiation is the process of finding the derivative of a function. In calculus, the derivative represents how a function changes as its input changes—essentially, it's the rate of change or slope of the function at any given point.
In the context of verifying integrals, differentiation plays a vital role. After evaluating an integral, like \( 2e^{\sqrt{y}} + C \) from the exercise, differentiating it helps confirm correctness:
In the context of verifying integrals, differentiation plays a vital role. After evaluating an integral, like \( 2e^{\sqrt{y}} + C \) from the exercise, differentiating it helps confirm correctness:
- You take the derivative of the result and simplify.
- If this newly differentiated expression matches the original integrand, then the integral was evaluated correctly.
- This step ensures accuracy by showing the reverse process, moving from the integral back to the original function.