Animated Solution for Mathematics - Definite Integration: The value 9∫09[x+110x]dx, where [t] denotes the greatest integer less than or equal to t, is _____
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Visualized Solution
Understanding the Integral I
Given integral: I=9∫09[x+110x]dx
[t] denotes the Greatest Integer Function (GIF).
Strategy: Break the integral at points where the inner function becomes an integer.
Defining the Inner Function f(x)
Let f(x)=x+110x
At lower limit x=0, f(0)=0
At upper limit x=9, f(9)=1090=3
The function increases from 0 to 3.
Finding Critical Point: f(x)=1
We need to find where f(x) crosses integer values.
Set f(x)=1⟹x+110x=1
Squaring both sides: x+110x=1
Solving for First Breakpoint x1
10x=x+1
9x=1⟹x=91
So, for x∈[0,91), [f(x)]=0.
Finding Critical Point: f(x)=2
Next integer is 2.
Set f(x)=2⟹x+110x=2
Squaring both sides: x+110x=4
Solving for Second Breakpoint x2
10x=4(x+1)
10x=4x+4
6x=4⟹x=32
So, for x∈[91,32), [f(x)]=1.
The Upper Limit x=9
We already know at x=9, f(9)=3.
So, for x∈[32,9), [f(x)]=2.
The integral breaks into three distinct regions.
Breaking the Integral I
I=9[∫01/90dx+∫1/92/31dx+∫2/392dx]
The first integral evaluates to zero.
Evaluating the Intervals
I=9[0+[x]1/92/3+[2x]2/39]
I=9[(32−91)+(18−34)]
Arithmetic Simplification
Simplify the first bracket: 32−91=96−1=95
Simplify the second bracket: 18−34=354−4=350
I=9[95+350]
Final Calculation of I
Distribute the 9:
I=(9×95)+(9×350)
I=5+3×50
I=5+150=155
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The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals
Solution Diagram
Analyzing the Setup
The Greatest Integer Function (GIF), denoted by [...], acts like a staircase. Instead of attempting a complex integration, we identify the points where the function f(x)=x+110x jumps between integer values.
As x increases from 0 to 9, the function f(x) climbs monotonically from f(0)=0 to f(9)=3. Because the function is continuous and increasing, it must pass through the integer values 1 and 2.
Finding the Jump Points
To determine the intervals where the GIF remains constant, we solve for the values of x where the function hits these integers.
For the first jump, we set the function equal to 1:
x+110x=1
Squaring both sides yields:
x+110x=1⇒10x=x+1⇒9x=1⇒x=91
For the second jump, we set the function equal to 2:
x+110x=2
Squaring both sides yields:
x+110x=4⇒10x=4x+4⇒6x=4⇒x=32
Evaluating the Integral
We now partition the interval [0,9] based on these jump points to evaluate the integral I=9∫09[x+110x]dx:
1. On the interval [0,1/9], the function value is between 0 and 1, so the GIF is 0.
2. On the interval [1/9,2/3], the function value is between 1 and 2, so the GIF is 1.
3. On the interval [2/3,9], the function value is between 2 and 3, so the GIF is 2.
The integral becomes the sum of three distinct areas:
I=9(∫01/90dx+∫1/92/31dx+∫2/392dx)
Final Calculation
Calculating the lengths of these intervals:
- The first part is 0.
- The second part is 1×(32−91)=96−1=95.
- The third part is 2×(9−32)=2×(327−2)=2×325=350.
Summing these values and multiplying by the external factor of 9: