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Evaluate to find f(x): a. \int 6x-7 dx b. \int cos (4x) dx c. \int sec^2(x) -...

Question:

Evaluate to find {eq}f(x){/eq}:

a. {eq}\int 6x-7 dx {/eq}

b. {eq}\int cos (4x) dx {/eq}

c. {eq}\int sec^2(x) - \frac{2}{\sqrt[3]{2}} {/eq}

d. {eq}\int \frac{4}{ \sqrt{1 -x^2}} dx {/eq}

e. {eq}\int 3x^2 + x - 5 dx {/eq}

f. {eq}\int 86x + e^{-4x} dx {/eq}

g. {eq}\int 8e^x + 10 e^{5x} dx{/eq}

Integrals:

In order to evaluate the given integrals, we will make use of the basic rules and standard formulae of the integral calculus. The formulae used here are given as: {eq}\begin{align*} \int x^n \ dx = \frac{x^{n+1}}{n+1} + c \ \ ; \ \ \int e^{nx} \ dx = \frac{e^{nx}}{\frac{d}{dx}(nx)} + c \ \ ; \ \ \int \cos x\ dx = \sin x + c \end{align*} {/eq}.

Answer and Explanation:


a. {eq}\int 6x-7 dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \frac{ 6x^2}{2}-7x + c \\ \\ \ & = 3x^2-7x + c \end{align*} {/eq}


b. {eq}\int cos (4x) dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \frac{\sin 4x}{\frac{d}{dx}(4x)} \\ \\ \ & = \frac{\sin 4x}{4}+ c \end{align*} {/eq}


c. {eq}\int sec^2(x) - \frac{2}{\sqrt[3]{2}} {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \tan x - \frac{2x}{\sqrt[3]{2}} + c \\ \\ \ & = \tan x - \frac{x}{\sqrt2} + c \end{align*} {/eq}


d. {eq}\int \frac{4}{ \sqrt{1 -x^2}} dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & =4 \int \frac{1}{ \sqrt{1 -x^2}} dx \\ \\ \ & =4 \sin^{-1}x + c \ \ \ \ \ \ \ \ \ \ \ \ \left[ \int \frac{1}{ \sqrt{1 -x^2}} \ dx= \sin^{-1}x + c\right] \end{align*} {/eq}


e. {eq}\int 3x^2 + x - 5 dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \frac{3x^3}{3}+ \frac{x^2}{2}- 5x+ c \\ \\ \ & = x^3+ \frac{x^2}{2}- 5x+ c \end{align*} {/eq}


f. {eq}\int 86x + e^{-4x} dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \frac{86x^2}{2}+ \frac{e^{-4x}}{\frac{d}{dx}(-4x)} \\ \\ \ & = 43x^2 - \frac{e^{-4x}}{4} + c \end{align*} {/eq}


g. {eq}\int 8e^x + 10 e^{5x} dx {/eq}

Integrating the above, we get:

{eq}\begin{align*} \ & = \frac{8e^{x}}{\frac{d}{dx}(x)} + \frac{10e^{5x}}{\frac{d}{dx}(5x)} \\ \\ \ & = \frac{8e^{x}}{1} + \frac{10e^{5x}}{5}+ c \\ \\ \ & = 8e^{x}+ 2e^{5x}+ c \end{align*} {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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