Evaluate the definite integral: \int_3^4 (\frac{3}{x^2}-1)dx

Question:

Evaluate the definite integral: {eq}\int_3^4 (\frac{3}{x^2}-1)dx{/eq}

Integration:

Let us consider a function f(t) where t is time, then the integral to that function is found out as follows:

{eq}\displaystyle \int f(t) \, dt = F(t) + C {/eq}.

From the above equation, {eq}F(t) {/eq} is the integral and C is an arbitrary constant.

Sum Rule:

If the functions given are {eq}f(x) , g(x) {/eq} then

{eq}\displaystyle \int f(x) + g(x) dx = \int f(x)dx + \int g(x) dx {/eq}

Formulas used:

{eq}\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} {/eq}

Answer and Explanation:

Given function is

{eq}\displaystyle \int_3^4 (\frac{3}{x^2} - 1)dx {/eq}

By using sum rule above equation can be written as

{eq}\displaystyle =\int_3^4 \frac{3}{x^2}dx - \int_3^4 1 dx {/eq}

{eq}\displaystyle =\left[\frac{-3}{x}\right]_3^4 - \left[x\right]_3^4 {/eq}

{eq}\displaystyle =(\frac{-3}{4}-\frac{-3}{3}) - (4-3) {/eq}

{eq}\displaystyle =\frac{-3}{4}+1-1 {/eq}

{eq}\displaystyle =\frac{-3}{4} {/eq}

The answer is

{eq}\displaystyle \int_3^4 (\frac{3}{x^2} - 1) dx=\frac{-3}{4} {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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