# 1. Find the most general antiderivative. (a) \ \int(10x^4 + 15x^2 - 7) dx \\ (b) \ \int (7cos \ x...

## Question:

1. Find the most general antiderivative.

{eq}(a) \ \int(10x^4 + 15x^2 - 7) dx \\ (b) \ \int (7cos \ x + \frac{3}{x}) dx \\ (c) \ \int(6 \cdot \sqrt{x} - \frac{15}{x^4} ) dx {/eq}

## Indefinite Integral:

An integration {eq}\displaystyle \int f(x) dx {/eq}, where upper and lower limit is not defined is called an indefinite integral.

For the above problem we use some of the following formulas:

(1) {eq}\displaystyle \int x^n dx=\dfrac{x^{(n+1)}}{n+1}+c {/eq}

(2) {eq}\displaystyle \int \cos x dx=\sin x+c {/eq}

(3) {eq}\displaystyle \int \dfrac{1}{x} dx=\ln (x)+c {/eq}.

Here {eq}\displaystyle c {/eq} is an integrating constant.

(a) {eq}\displaystyle \int(10x^4 + 15x^2 - 7) dx =10.\dfrac{x^5}{5}+15\dfrac{x^3}{3}-7x+c =\color{blue}{2x^5+5x^3-7x+c} {/eq}.

(b) {eq}\displaystyle \int (7\cos \ x + \dfrac{3}{x}) dx =\color{blue}{ 7\sin x+3\ln x+c} {/eq}.

(c) {eq}\displaystyle \int(6 \cdot \sqrt{x} - \frac{15}{x^4} ) dx =\int (6 x^{\dfrac{1}{2}}-15x^{(-4})) dx =6\times \dfrac{x^{1+\dfrac{1}{2}}}{\dfrac{1}{2}+1}-15\dfrac{x^{-4+1}}{(-4+1)}+c =6\times \dfrac{2}{3}x^{\dfrac{3}{2}}+\dfrac{15}{3} x^{-3}+c =\color{blue}{4x^{\dfrac{3}{2}}+\dfrac{5}{x^3}+c} {/eq}.

Here {eq}\displaystyle c {/eq} is an integrating constant.

Work as an Integral

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Chapter 7 / Lesson 9
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After watching this video, you will be able to solve calculus problems involving work and explain how that relates to the area under a force-displacement graph. A short quiz will follow.