Question

Constant of Integration Consider the integral $$\int \sqrt{x+1} d x$$ (a) Show that \(\int \sqrt{x+1} d x=\frac{2}{3}(x+1)^{3 / 2}+C\) (b) Writing to Learn Explain why $$y_{1}=\int_{0}^{x} \sqrt{t+1} d t\( and \)y_{2}=\int_{3}^{x} \sqrt{t+1} d t$$ are antiderivatives of \(\sqrt{x+1}\) (c) Use a table of values for \(y_{1}-y_{2}\) to find the value of \(C\) for which \(y_{1}=y_{2}+C\) (d) Writing to Learn Give a convincing argument that $$C=\int_{0}^{3} \sqrt{x+1} d x$$

Short answer

The integral \(\int \sqrt{x+1} dx = 2/3 (x+1)^{3/2} + C\) can be confirmed by applying the integral power rule. The functions \(y_1\), \(y_2\) are antiderivatives of \(\sqrt{x+1}\) as they can be expressed in the form \(F(x) - F(a)\), made possible by the Fundamental Theorem of Calculus. The constant of integration, C, is determined to be \(14/3\) as this is the difference between the antiderivatives \(y_1\) and \(y_2\) for the defined range. A convincing argument that C equals \(\int_{0}^{3} \sqrt{x+1} dx\), is established by direct comparison of the evaluated expressions.

Step-by-step solution

Confirm the Integral

We first re-write \( \int \sqrt{x+1} d x\) as \(\int (x+1)^{1/2} d x\) and perform the integral using the power rule, which states \(\int x^n dx = 1/(n+1) x^{n+1} + C\), where n is any real number. Applying the power rule, we obtain \((1/2 + 1) \cdot (1 / (1/2 + 1))(x+1)^{1/2 + 1} + C\), which simplifies to \(2/3 (x+1)^{3/2} + C\). This confirms the equation in (a).

Explain Antiderivatives

The definitions of \(y_1\) and \(y_2\) both involve integrals of \(\sqrt{t+1}\) over an interval ending in x. Since the Fundamental Theorem of Calculus states the integral of a derivative of a function over an interval [a,x] is simply that function evaluated at x minus that function evaluated at a, both \(y_1\) and \(y_2\) can be expressed as \(F(x) - F(a)\), where F is any antiderivative of \(\sqrt{t+1}\), thus \(y_1\) and \(y_2\) are antiderivatives of \(\sqrt{x+1}\).

Determine the Constant of Integration

Here, we compute \(y_1 - y_2\) for a few values of x. As x varies, we find that \(y_1 - y_2\) equals \(\int_{0}^{3} \sqrt{t+1} dt\), and we conclude that \(y_1 - y_2 = C\).

Integrate to Determine C

To give a convincing argument that C equals \(\int_{0}^{3} \sqrt{x+1} dx\), it is enough to show their values are equal. We already argued in Step 3 that \(C = y_1 - y_2 = \int_{0}^{3} \sqrt{t+1} dt\). By directly computing, we see that \(\int_{0}^{3} \sqrt{x+1} dx\) also equals \(2/3 (3+1)^{3/2} - 2/3 (0+1)^{3/2} = 16/3 - 2/3 = 14/3\), which matches the value we computed for C in Step 3.

Key concepts

Constant of Integration

In integration calculus, the constant of integration is a fundamental concept. Whenever we find an antiderivative or an integral of a function, we are finding all possible functions that could differentiate to give us the original function. However, because differentiation wipes out any constant part of a function (since the derivative of a constant is zero), we add a 'C' to our integral to represent any possible constant that was lost during differentiation.

Through a problem, such as finding \( \int \sqrt{x+1} \, dx \) and arriving at the solution \( \frac{2}{3}(x+1)^{3 / 2}+C \), what we're saying is 'I've found all functions whose derivative is \( \sqrt{x+1} \), and they differ by some constant C.' This C could be any real number and it makes the family of antiderivatives complete. Understanding this concept helps students see why different definite integrals of the same function over different intervals can lead to different values, all of which relate to the 'C' value in the indefinite integral.

Antiderivatives

An antiderivative of a function is another function that, when differentiated, gives back the original function. For instance, regarding the function \( \sqrt{x+1} \), any function that differentiates to \( \sqrt{x+1} \) is called its antiderivative. These are not limited to a single function due to the constant of integration, as discussed previously.

When we talk about another related integral, such as \( y_{1} = \int_{0}^{x} \sqrt{t+1} \, dt \) or \( y_{2} = \int_{3}^{x} \sqrt{t+1} \, dt \) (from the textbook problem), we are actually looking at specific antiderivatives of \( \sqrt{x+1} \) that pass through the points (0,0) and (3,0) on their respective graphs. This illustrates how two different antiderivatives of the same function can differ by a constant. Comparing these two integrals shows the constant of integration in action and helps students understand the infinite set of antiderivative functions that exist for a given function.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key player in connecting differentiation and integration, the two main concepts in calculus. This theorem has two parts. The first part states that if 'F' is an antiderivative of a function 'f' over an interval [a, b], then the integral of 'f' from 'a' to 'b' is given by the difference F(b) - F(a).

With an example like \( y_{1} = \int_{0}^{x} \sqrt{t+1} \, dt \), the theorem tells us that we are essentially finding the function value at 'x' minus the function value at '0', showing us the net change in the function between these two points. In our textbook problem, this theorem underlines the fact that \( y_{1} \) and \( y_{2} \) are fundamentally the same function, expressed over different intervals, and therefore must be related by a constant. Understanding this theorem allows students to grasp the seamless transition from the concept of accumulation (integration) to the concept of rate of change (differentiation), empowering them to solve complex calculus problems with deeper insight.

Power Rule for Integration

The power rule for integration is a straightforward technique that students can use to find the integral of powers of 'x'. The rule is \( \int x^n \, dx = \frac{1}{n+1} x^{n+1} + C \), where 'n' is a real number besides -1. When applying this rule, students must remember to increase the exponent by one and divide by the new exponent, and then, do not forget to add the constant of integration!

For the given exercise's integral, \( \int \sqrt{x+1} \, dx \), the power rule is used after rewriting \( \sqrt{x+1} \) as \( (x+1)^{1/2} \). Following the power rule, we increase the exponent to \( 3/2 \) and multiply by the reciprocal of this exponent to get \( \frac{2}{3}(x+1)^{3/2} + C \). This application is integral (pun intended) to calculating integrals of polynomials and helps students simplify complex antiderivative problems. The power rule is an essential tool in any student's calculus toolkit, paving the way for tackling more advanced integration techniques.