Animated Solution for Mathematics - Definite Integration: Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral ∫01[−8x2+6x−1]dx is equal to
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Visualized Solution
Understanding the Problem
Let f(x)=−8x2+6x−1
The integral is I=∫01[f(x)]dx, where [⋅] is the greatest integer function.
Goal: Identify intervals where [f(x)] is constant.
Analyzing the Function f(x)
Vertex: x=−2ab=−2(−8)6=83
Max value: f(83)=81
Endpoints: f(0)=−1 and f(1)=−3
Range of f(x) for x∈[0,1] is [−3,81]
Finding Roots: f(x)=0
Set f(x)=0⇒−8x2+6x−1=0
(4x−1)(2x−1)=0
Roots: x=41 and x=21
For x∈[41,21], f(x)∈[0,81], so [f(x)]=0
Finding Points: f(x)=−1
Set f(x)=−1⇒−8x2+6x−1=−1
−8x2+6x=0⇒2x(3−4x)=0
Points: x=0 and x=43
For x∈[0,41) and x∈(21,43], [f(x)]=−1
Finding Points: f(x)=−2
Set f(x)=−2⇒−8x2+6x−1=−2
8x2−6x−1=0
x=166±36−4(8)(−1)=83±17
For x∈(43,83+17], [f(x)]=−2
Finding Points: f(x)=−3
The remaining interval is up to x=1
At x=1, f(1)=−3
For x∈(83+17,1], [f(x)]=−3
Splitting the Integral
I=∫041(−1)dx+∫4121(0)dx+∫2143(−1)dx
+∫4383+17(−2)dx+∫83+171(−3)dx
Calculating the First Three Parts
I1=−1⋅(41−0)=−41
I2=0⋅(21−41)=0
I3=−1⋅(43−21)=−41
Sum of first three =−41+0−41=−21
Calculating the Fourth Part
I4=−2⋅(83+17−43)
I4=−2⋅(83+17−6)
I4=−2⋅(817−3)=86−217
Calculating the Fifth Part
I5=−3⋅(1−83+17)
I5=−3⋅(88−3−17)
I5=−3⋅(85−17)=8317−15
Final Summation and Answer
I=−21+86−217+8317−15
I=8−4+6−217+317−15
I=817−13
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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals
Solution Diagram
Analyzing the Setup
Imagine you are standing before a complex, shifting landscape. You have a quadratic function, f(x)=−8x2+6x−1, and you are asked to find the area under its 'staircase' version, the greatest integer function [f(x)].
When we see [f(x)], we shouldn't panic. Instead, we should see a staircase where f(x) is the smooth, continuous path and the greatest integer function is the rigid, stepped structure that follows it.
Our goal is to calculate the area under this staircase from x=0 to x=1.
Mapping the Terrain
First, let us understand the behavior of our quadratic function f(x)=−8x2+6x−1. This is an inverted parabola.
To understand its journey, we find its vertex using:
x=−2ab=−2(−8)6=83
At this peak, the value is f(83)=81. At the boundaries, f(0)=−1 and f(1)=−3. This tells us that as x moves from 0 to 1, the function climbs from −1 to 1/8 and then descends to −3.
Finding the Jump Points
The greatest integer function [f(x)] only changes its value when f(x) crosses an integer. We need to find exactly where this happens.
We set f(x)=0, which gives us (4x−1)(2x−1)=0, yielding roots at x=1/4 and x=1/2. In this interval, f(x) is between 0 and 1/8, so [f(x)]=0.
Next, we look for f(x)=−1. Solving −8x2+6x−1=−1 gives x=0 and x=3/4.
Finally, for f(x)=−2, we solve −8x2+6x−1=−2, or 8x2−6x−1=0. Using the quadratic formula, we find:
x=83±17
We take the positive root for the right side of the parabola.
The Sum of Rectangles
Now, the integral I=∫01[f(x)]dx transforms into a sum of simple rectangular areas:
Each integral is just the height of the step multiplied by the width of the interval. The first three parts are straightforward: I1=−1/4, I2=0, and I3=−1/4. Summing these gives −1/2.
For the fourth part:
I4=−2⋅(83+17−43)=86−217
For the final part:
I5=−3⋅(1−83+17)=8317−15
Final Synthesis
Adding these together:
I=−21+86−217+8317−15
Converting −1/2 to −4/8, we get:
I=8−4+6−217+317−15=817−13
This is the beauty of calculus. We took a complex, discontinuous function and, by understanding its geometric soul, broke it down into simple, manageable pieces. You have successfully navigated the staircase!