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Find the indefinite integral, and check your answer by differentiation. $$ \int\left(2 x^{9}+3 e^{x}+4\right) d x $$

Short Answer

Expert verified
The indefinite integral of the given function is: \[ \int (2x^9 + 3e^x + 4) dx = \frac{x^{10}}{5} + 3e^x + 4x + C \] This was verified by differentiating the result and obtaining the initial function: \[ 2x^9 + 3e^x + 4 \]

Step by step solution

01

Find the indefinite integral of the first term (polynomial term)

To find the indefinite integral of the polynomial term \(2x^9\), we can use the power rule for integration, which states that \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\), where n is a constant. In this case, \(n=9\), so the indefinite integral of \(2x^9\) is: \[ \int 2x^9 dx = 2\int x^9 dx = 2\frac{x^{9+1}}{9+1}+C_1 = \frac{x^{10}}{5}+C_1 \]
02

Find the indefinite integral of the second term (exponential term)

To find the indefinite integral of the exponential term \(3e^x\), we recall that the integral of \(e^x\) is simply \(e^x\). Therefore, the indefinite integral of \(3e^x\) is: \[ \int 3e^x dx = 3\int e^x dx = 3e^x + C_2 \]
03

Find the indefinite integral of the third term (constant term)

To find the indefinite integral of the constant term 4, we recall that the integral of a constant is the constant times the variable. Therefore, the indefinite integral of 4 is: \[ \int 4 dx = 4x + C_3 \]
04

Combine the indefinite integrals of the three terms

Now that we have found the indefinite integrals of each of the three terms, we can combine them to find the indefinite integral of the entire function: \[ \int (2x^9 + 3e^x + 4) dx = \frac{x^{10}}{5} + C_1 + 3e^x + C_2 + 4x + C_3 \] Combining the constants of integration \(C_1\), \(C_2\), and \(C_3\) into a single constant \(C\), we get: \[ \int (2x^9 + 3e^x + 4) dx = \frac{x^{10}}{5} + 3e^x + 4x + C \]
05

Check the answer by differentiation

To check if our answer is correct, we will differentiate it and see if we get the original function. The derivative of our integrated function is: \[ \frac{d}{dx}\left(\frac{x^{10}}{5} + 3e^x + 4x + C\right) = \frac{10x^9}{5} + 3e^x + 4 \] Simplifying the first term: \[ \frac{10x^9}{5} = 2x^9 \] Thus, the derivative is: \[ 2x^9 + 3e^x + 4 \] Since the derivative matches the original function, our indefinite integral is correct: \[ \int (2x^9 + 3e^x + 4) dx = \frac{x^{10}}{5} + 3e^x + 4x + C \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration Rules
Understanding integration rules is crucial for calculating indefinite integrals, which represent the antiderivative of a function. When working with indefinite integrals, you'll often encounter various types of functions, such as polynomials and exponential functions, each with their specific rules for integration. Let's break down these rules for easier comprehension.

Firstly, when integrating a polynomial expression like a constant or a power of a variable, you add one to the exponent and divide by the new exponent, then attach a constant of integration 'C'. This comes from the power rule. For example, integrating a constant 'a' gives us 'ax + C'.

Additionally, when multiple terms are involved, as in the exercise provided, you integrate each term separately. This aligns with the sum rule of integration that allows us to break down complex integrals into simpler parts. Lastly, when dealing with a constant multiplied by a function, like in the expression '2x^9', you can first pull out the constant and integrate the remaining function. These rules simplify the integration process, enabling a step-by-step approach to solving complex indefinite integrals.
Power Rule for Integration
The power rule for integration is one of the fundamental techniques for integrating polynomial expressions. It's elegantly simple and incredibly useful. Under the power rule, the integral of 'x' raised to the power 'n', where 'n' is any real number except -1, is given by \( \frac{x^{n+1}}{n+1} + C \). The constant 'C' represents the constant of integration which arises because the integral is indefinite.

Using the power rule, the integral of \( 2x^9 \) from our exercise yields \( \frac{2}{10}x^{10} + C \) or simplified to \( \frac{x^{10}}{5} + C \) after multiplying by the constant 2. This is a clear demonstration of the power rule's application. Remember, this rule does not apply when 'n' equals -1 because the integral of \( \frac{1}{x} \) is the natural logarithm of '|x|', not a rational function.
Exponential Functions
Exponential functions are another class of functions that often appear in calculus problems. The most common base for exponential functions in calculus problems is 'e', also known as Euler's number, approximately equal to 2.71828. The indefinite integral of \( e^x \) is special because it is one of the few functions that remains unchanged upon integration.

Thus, the integral of \( e^x \) is simply \( e^x + C \) where 'C' is the constant of integration. In our exercise, we integrate \( 3e^x \) and land at \( 3e^x + C \), as multiplying by the constant 3 comes before the integration. It is important to note that this property holds only for the natural exponential function, not for other bases, which require a different integration approach.

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Most popular questions from this chapter

If \(f(x)=\int_{2}^{x} \frac{d t}{\sqrt{1+t^{3}}}\), where \(x>-1\), what is \(\left(f^{-1}\right)^{\prime}(0)\) ?

a. Show that \(\int_{0}^{\pi} x f(\sin x) d x=(\pi / 2) \int_{0}^{\pi} f(\sin x) d x\). Hint: Use the substitution \(x=\pi-u\). b. Use the result of part (a) to evaluate \(\int_{0}^{\pi} x \sin x d x\).

Special Theory of Relativity According to Einstein's special theory of relativity, the mass of a particle is given by $$ m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ where \(m_{0}\) is the rest mass of the particle, \(v\) is its velocity, and \(c\) is the speed of light. Suppose that a particle starts from rest at \(t=0\) and moves along a straight line under the action of a constant force \(F\). Then, according to Newton's second law of motion, the equation of motion is $$ F=m_{0} \frac{d}{d t}\left(\frac{v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right) $$ Find the velocity and position functions of the particle. What happens to the velocity of the particle as time goes by?

Express the area of the region under the graph of the function f over the interval as the limit of a sum (use the right endpoints), (b) use a computer algebra system (CAS) to find the sum obtained in part (a) in compact form, and (c) evaluate the limit of the sum found in part (b) to obtain the exact area of the region. $$ f(x)=x^{5} ; \quad[0,2] $$

A car travels along a straight road in such a way that the average velocity over any time interval \([a, b]\) is equal to the average of its velocities at \(a\) and at \(b\). a. Show that its velocity \(v(t)\) satisfies $$\int_{a}^{b} v(t) d t=\frac{1}{2}[v(a)+v(b)](b-a)$$ b. Show that \(v(t)=c t+d\) for some constants \(c\) and \(d\). Hint: Differentiate Equation (1) with respect to \(a\) and with respect to \(b\).

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