# Compute \int_{0}^{3} (\frac{1}{2}x -1)dx a) the fundamental theorem of calculus b) by...

## Question:

Compute {eq}\int_{0}^{3} (\frac{1}{2}x -1)dx{/eq}

a) the fundamental theorem of calculus

b) by interpreting the integral in terms of area

## The Fundamental Theorem of Calculus:

If {eq}f {/eq} is continuous on {eq}\left[a,b\right] {/eq} and {eq}F {/eq} is any antiderivative of {eq}f {/eq} on {eq}\left[a,b\right] {/eq}, then

{eq}\displaystyle \int_{a}^{b}f(x)\:dx=F(b)-F(a)\\ {/eq}.

The Definite Integral as Area Under a Curve:

If {eq}f(x) {/eq} is a continuous function for {eq}b \geq a {/eq}, the area of the region bounded by the function and {eq}x-axis {/eq} on the interval {eq}\left[a,b\right] {/eq} is

{eq}Area=\int_{a}^{b}f(x)dx {/eq}

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Step 1. Evaluate {eq}\int_{0}^{3} (\frac{1}{2}x -1)dx {/eq} utilizing the fundamental theorem of calculus.

{eq}\begin{align} \displaystyle...

The Fundamental Theorem of Calculus

from

Chapter 12 / Lesson 10
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The fundamental theorem of calculus links derivatives and antiderivatives in order to find the area under a curve. Learn more about the theorem with an example using velocity.