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Chapter 7: Problem 10

Evaluate the following integrals. $$\int e^{3-4 x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral: \(\int e^{3-4x} dx\) Answer: The integral evaluates to \(-\frac{1}{4} e^{3-4x} + C\), where \(C\) is the constant of integration.

Step by step solution

01

Identify the substitution

To integrate the given function, we will substitute a variable for the term inside the exponential function. Let's choose \(u = 3 - 4x\). Then, we need to find the derivative of \(u\) with respect to \(x\), which we'll represent as \(du/dx\).
02

Find the derivative of the substitution variable

Find the derivative of \(u\) with respect to \(x\): $$\frac{du}{dx} = \frac{d}{dx}(3 - 4x)$$ $$\frac{du}{dx} = -4$$ We must also express \(dx\) in terms of \(du\), so we can replace it in the integral. Since \(\frac{du}{dx}=-4\), this means \(dx = \frac{du}{-4}\).
03

Substitute the variable and integrate

Now let's substitute \(u\) and \(dx\) into the integral: $$\int e^{3-4x}dx = \int e^u \frac{du}{-4}$$ Now we can take the constant \(-1/4\) out of the integral and integrate the exponential function: $$-\frac{1}{4}\int e^u du$$ The integral of the exponential function \(e^u\) is just \(e^u\). Therefore, we have: $$-\frac{1}{4} e^u + C$$ where \(C\) is the constant of integration.
04

Substitute the original variable back

Now, we need to express our result in terms of the original variable \(x\). Recall the substitution we made: \(u=3-4x\). Substitute \(u\) back into the result: $$-\frac{1}{4} e^{3-4x} + C$$ So the final solution for the given integral is: $$\int e^{3-4x} dx = -\frac{1}{4} e^{3-4x} + C$$

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Most popular questions from this chapter

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

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Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$

Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.
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