Question

Evaluate the following integrals. $$\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x$$

Short answer

The value of the definite integral $\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x$ is $\frac{4}{\ln 2}$.

Step-by-step solution

Identify the substitution

We will use the substitution method to simplify the integral. Let us define a new variable, \(u\), as: $$u = \sqrt{x}$$ Now, we need to find the corresponding differential \(du\) in terms of \(dx\).

Find the differential

Take the derivative of \(u\) with respect to \(x\): $$\frac{d u}{d x} = \frac{1}{2\sqrt{x}}$$ Now, multiply both sides by \(dx\): $$du = \frac{1}{2\sqrt{x}} dx$$ We can now rewrite our integral in terms of the new variable \(u\).

Rewrite the integral

Using the substitution we made, we can rewrite the integral as follows: $$\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x = \int \frac{2^u}{\sqrt{x}} \cdot 2\sqrt{x} du$$ Observe that \(\frac{2^u}{\sqrt{x}} \cdot 2\sqrt{x}\) simplifies to \(2^{u+1}\). Additionally, we need to change the limits of integration. When \(x = 1\), \(u = \sqrt{1} = 1\), and when \(x = 4\), \(u = \sqrt{4} = 2\). Thus, the new limits of integration are \(1\) and \(2\). Our integral becomes: $$\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x = \int_{1}^{2} 2^{u+1} d u$$

Evaluate the integral

Now, we can integrate: $$\int_{1}^{2} 2^{u+1} d u = \int_{1}^{2} 2 \cdot 2^u d u$$ To integrate the expression \(2^u\), we use the formula: $$\int a^u d u = \frac{a^u}{\ln a} + C$$ So, we have: $$\int_{1}^{2} 2 \cdot 2^u d u = 2 \int_{1}^{2} 2^u d u = 2 \left( \frac{2^u}{\ln 2} \right)\Big|_{1}^{2}$$

Compute the definite integral

Finally, we can compute the value of the definite integral: $$2\left(\frac{2^2}{\ln 2} - \frac{2^1}{\ln 2}\right) = 2\left(\frac{4 - 2}{\ln 2}\right) = 2\left(\frac{2}{\ln 2}\right) = \frac{4}{\ln 2}$$ Thus, the value of the integral is: $$\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x = \frac{4}{\ln 2}$$

Key concepts

Integration

Integration is one of the two core operations of calculus, with differentiation being the other. It is essentially the reverse process of differentiation and is used to find accumulations, such as areas under curves, volumes, and more general accumulations of quantities.
In the context of definite integrals, like in the original problem, integration calculates the net area between the curve and the x-axis over a specified interval.

• When integrating, you're summing up an infinite number of infinitesimally small rectangles under the curve.
• This calculation provides a value representing the total accumulation between the limits of integration.

Integrals can be evaluated

• definitely, where the limits of integration are specified, providing a numerical result,
• or indefinitely, providing a general form solution.

Understanding integration is crucial for solving complex problems involving cumulative quantities effectively.

Substitution Method

The substitution method is a powerful tool in calculus used to simplify the process of integration. It revolves around changing variables to make an integral easier to solve. Think of it as a 'change of perspective', where you replace a complex variable with a simpler one.
Here's how it works:

• Select a new variable, typically denoted as \(u\), that represents a part of the original function.
• Find the differential \(du\) in terms of \(dx\) by taking the derivative of \(u\).
• Rewrite the integral in terms of the new variable \(u\), adjusting the limits of integration if necessary.

This method is particularly useful when dealing with composite functions or when the original integral features complex algebraic, trigonometric, or exponential expressions.
Substitution transforms the integral into a form that is easier to manage, turning a potentially overwhelming problem into one that's much more straightforward.

Definite Integral

A definite integral is an integral with specific upper and lower limits. Unlike indefinite integrals, definite integrals provide a numeric value, which represents the total accumulation over a particular interval. These limits confine the area calculation to partial regions of the function's curve above the x-axis within the set range.
The main components of evaluating a definite integral include:

• Integrating the function across the defined limits.
• Subtracting the value of the integral at the lower limit from the value at the upper limit.
• Considering negative values when the curve is below the x-axis, leading to net area calculations which account for these segments as well.

Definite integrals are ubiquitous in calculus, helping to solve problems from physics, engineering, economics, and more, providing solutions to complex real-world problems by calculating precise quantities.