Question

Find the area under \(y=1 / x\) and above the \(x\) -axis from \(x=1\) to \(x=4\) $$ \frac{d}{d x} \ln \left(x+\sqrt{x^{2}+1}\right)=\frac{1}{\sqrt{1+x^{2}}} $$

Short answer

The area is \( \ln(4) \) or approximately 1.3863.

Step-by-step solution

Understand the Problem

We are asked to find the area under the curve \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 4 \). This can be achieved by calculating the definite integral of \( \frac{1}{x} \) over the interval \([1, 4]\).

Set Up the Integral

The integral to find the area under \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 4 \) is:\[\int_{1}^{4} \frac{1}{x} \, dx\]

Find the Antiderivative

The antiderivative of \( \frac{1}{x} \) is \( \ln|x| \). So, the antiderivative of our function is:\[\ln|x|\]

Evaluate the Definite Integral

Substitute the bounds into the antiderivative:\[\left[ \ln|x| \right]_{1}^{4} = \ln(4) - \ln(1)\]Since \( \ln(1) = 0 \), it simplifies to:\[\ln(4)\]

Final Calculation

Calculate \( \ln(4) \). If needed, use an approximation. \( \ln(4) \approx 1.3863 \). This gives the area under the curve from \( x = 1 \) to \( x = 4 \).

Key concepts

Antiderivative

Understanding the concept of an antiderivative is crucial when dealing with integrals. An antiderivative, sometimes known as an indefinite integral, is a function that reverses the process of differentiation. If you have a function like \( f(x) \), an antiderivative of \( f \) is a function, let's call it \( F(x) \), such that the derivative of \( F \) gives you back \( f(x) \). The process of finding an antiderivative is called integration, and it is essential for computing areas under curves. - For example, the antiderivative of \( \frac{1}{x} \) is \( \ln|x| \), because when you differentiate \( \ln|x| \), you get back \( \frac{1}{x} \).When dealing with definite integrals, which have specific limits (like from 1 to 4 in our example), you use the antiderivative to find the value of the integral between those limits. Remember that the antiderivative represents a family of functions, differing by a constant. However, when evaluating definite integrals, these constants cancel out, so they are often omitted.

Natural Logarithm

The natural logarithm is a fundamental concept in calculus and is denoted by \( \ln(x) \). It is the inverse of the exponential function with base \( e \), where \( e \approx 2.71828 \). The logarithm \( \ln(x) \) tells us the power to which \( e \) must be raised to obtain the number \( x \).- For instance, \( \ln(4) \) is the power to which \( e \) must be raised to get 4. This can also be expressed using logarithmic rules, such as \( \ln(ab) = \ln(a) + \ln(b) \).The natural logarithm has a unique property when dealing with integrals. Its derivative is \( \frac{1}{x} \), which means that finding the antiderivative of \( \frac{1}{x} \) results in the natural logarithm function, \( \ln|x| \). This makes it especially useful for problems involving areas under curves of functions like \( \frac{1}{x} \), as seen in our exercise. The natural logarithm simplifies complex algebraic expressions significantly and is an essential tool in calculus for solving real-world problems.

Area Under a Curve

Finding the area under a curve is a fundamental application of definite integrals in calculus. The area under the curve represents the total accumulation of a quantity, rather like how finding the area of a rectangle gives us its total space. To find the area under the curve of a function like \( y = \frac{1}{x} \) from \( x = 1 \) to \( x = 4 \), we use the definite integral \[\int_{1}^{4} \frac{1}{x} \, dx\]This integral effectively sums up the infinitely small slices of area between the curve and the x-axis from 1 to 4. - We first find the antiderivative, which is \( \ln|x| \).- Then, we evaluate this antiderivative at the bounds of our interval, which gives \( \ln(4) - \ln(1) \). - Since \( \ln(1) = 0 \), the calculation simplifies to \( \ln(4) \).Thus, the area under the curve from \( x = 1 \) to \( x = 4 \) is represented by \( \ln(4) \), providing a practical example of how calculus can measure areas that are not straightforward rectangles or triangles.