Problem 63
Question
Prove that $$\int_{a}^{b} x d x=\frac{b^{2}-a^{2}}{2}$$
Step-by-Step Solution
Verified Answer
Thus, \(∫_{a}^{b} x dx = \frac{b^{2}-a^{2}}{2}\).
1Step 1: Find the indefinite integral
The indefinite integral of \(x\) is \(∫x dx = \frac{x^2}{2} + C\), where \(C\) is the constant of integration.
2Step 2: Apply the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus states that \(∫_{a}^{b} f(x) dx = F(b) - F(a)\), where \(F(x)\) is the antiderivative of \(f(x)\). So, \(∫_{a}^{b}xdx = \frac{b^2}{2} - \frac{a^2}{2}\).
3Step 3: Simplify the result
By combining the fraction, the result is \(\frac{b^2 - a^2}{2}\).
Key Concepts
Definite IntegralsAntiderivativesIndefinite Integrals
Definite Integrals
Definite integrals are an essential concept in calculus that help calculate the area under a curve between two specific points on the x-axis, often referred to as the limits of integration. In our example, these limits are from point \(a\) to point \(b\).
This process essentially lets us sum an infinite number of small areas under the curve.
The result will often be dependent on the substitution of these limits into an antiderivative, a function closely related to indefinite integrals. The process leverages the power of the Fundamental Theorem of Calculus.
This process essentially lets us sum an infinite number of small areas under the curve.
- Definite integrals result in a numerical value.
- The limits of integration \(a\) and \(b\) are crucial.
- The value of the definite integral represents the net area between the function and the x-axis.
The result will often be dependent on the substitution of these limits into an antiderivative, a function closely related to indefinite integrals. The process leverages the power of the Fundamental Theorem of Calculus.
Antiderivatives
An antiderivative can be thought of as a function that 'undoes' the operation of differentiation. When you know a derivative \(f(x)\), finding its antiderivative \(F(x)\) means identifying the function of which it is the derivative. This step is crucial in integrating functions.
In simpler terms, if you differentiate the antiderivative, you get back the original function:
Identifying the correct antiderivative is key to applying definite integrals. The effort is made much clearer when utilizing the Fundamental Theorem of Calculus, which seamlessly connects antiderivatives to the evaluation of definite integrals.
In simpler terms, if you differentiate the antiderivative, you get back the original function:
- If \(f(x) = x\), then the antiderivative \(F(x) = \frac{x^2}{2}\) plus a constant \(C\).
- The constant \(C\) is introduced because differentiation of a constant is zero.
Identifying the correct antiderivative is key to applying definite integrals. The effort is made much clearer when utilizing the Fundamental Theorem of Calculus, which seamlessly connects antiderivatives to the evaluation of definite integrals.
Indefinite Integrals
While definite integrals provide a specific value, indefinite integrals involve the entire family of functions formed by adding varying constants to an antiderivative. This is because the integration process is essentially reversing differentiation, where the constant term is lost.
Mathematically, integrating \(f(x)=x\) results in the indefinite integral:
Understanding indefinite integrals is fundamental to acquiring the knowledge to find specific definite integrals, establishing a base for forming the antiderivative needed for precise area calculations between limits.
Mathematically, integrating \(f(x)=x\) results in the indefinite integral:
- \(∫x \, dx = \frac{x^2}{2} + C\)
- The \(+C\) represents an entire family of functions with different vertical shifts.
- Every different value of \(C\) gives you a unique function, which results from losing the constant during differentiation.
Understanding indefinite integrals is fundamental to acquiring the knowledge to find specific definite integrals, establishing a base for forming the antiderivative needed for precise area calculations between limits.
Other exercises in this chapter
Problem 63
In Exercises \(63-68,\) find the limit. \(\lim _{x \rightarrow \infty} \sinh x\)
View solution Problem 63
Verify each rule by differentiating. Let \(a>0\). $$ \int \frac{d u}{u \sqrt{u^{2}-a^{2}}}=\frac{1}{a} \operatorname{arcsec} \frac{|u|}{a}+C $$
View solution Problem 63
Find \(F\) as a function of \(x\) and evaluate it at \(x=2, x=5,\) and \(x=8\). $$ F(x)=\int_{1}^{x} \frac{10}{v^{2}} d v $$
View solution Problem 63
Show that the two formulas are equivalent. $$ \begin{array}{l} \int \tan x d x=-\ln |\cos x|+C \\ \int \tan x d x=\ln |\sec x|+C \end{array} $$
View solution