Calculus notes on properties of definite integrals with a graph of function f(x) and related problems.

Let's solve the problems step by step using the graph of $ f(x) $ and the properties of definite integrals.

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🔹 Graph Analysis

From the graph:
- The function $ f(x) $ is piecewise defined.
- We'll estimate areas under the curve geometrically (using shapes like triangles, rectangles, and semicircles).
- Areas above the x-axis are positive, and those below are negative.

We will compute definite integrals by breaking them into regions and calculating area.

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## ✔ Problem 1: $ \int_0^{12} f(x)\,dx $

This is the total signed area from $ x = 0 $ to $ x = 12 $. Break it into parts:

a. $ [0, 4] $: Semicircle-like shape

- Looks like a semicircle with radius 2 (from $ y=0 $ to $ y=2 $), centered at $ (2, 1) $? Wait — actually, let's analyze:
- From $ x=0 $ to $ x=4 $, the graph goes from $ (0,1) $ up to $ (2,2) $, down to $ (4,0) $
- This looks like a semicircle of radius 2, but only the upper half of a circle?
Wait — better to see it as a half-ellipse or just a semi-circular arc?

Actually, it's more likely a semicircle of radius 2 centered at $ (2,1) $, but that might be overcomplicating.

But notice: It starts at $ (0,1) $, peaks at $ (2,2) $, ends at $ (4,0) $. So this is symmetric about $ x=2 $, and forms a semicircle of radius 1? No — height is 1 unit above center?

Wait — let’s think differently.

Alternatively, observe:
- From $ x=0 $ to $ x=4 $, the curve is above the x-axis, and resembles a semicircle of radius 2, but shifted.

But looking closely:
- The curve from $ x=0 $ to $ x=4 $ goes from $ (0,1) $ to $ (2,2) $ to $ (4,0) $
- This is not a full semicircle.

Better approach: Break into known shapes.

Let’s divide the interval $[0,12]$ into segments based on changes in behavior:

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🔹 Breakdown of $ \int_0^{12} f(x)\,dx $

#### Interval [0,4]:
- Curve from $ (0,1) $ to $ (2,2) $ to $ (4,0) $
- This is a semicircle? Or a parabola? But since no formula, assume we can approximate.

But wait — look at the shape: symmetric around $ x=2 $, peak at $ (2,2) $, endpoints $ (0,1) $ and $ (4,0) $? Not symmetric.

Wait — maybe it's a quarter-circle?

No — better idea: use geometry.

Actually, from $ x=0 $ to $ x=4 $, the graph appears to be a semicircle of radius 2, but shifted?

Wait — let's check values:
- At $ x=0 $: $ f(0)=1 $
- At $ x=2 $: $ f(2)=2 $
- At $ x=4 $: $ f(4)=0 $

So this is not a perfect semicircle.

Alternative: Treat it as a combination of shapes

But perhaps it's easier to note that from $ x=0 $ to $ x=4 $, the curve is above the x-axis, and we can estimate area.

But wait — actually, look again: from $ x=0 $ to $ x=4 $, it looks like a semicircle of radius 2, but centered at (2,1)? Let’s test:

- Distance from $ (2,1) $ to $ (0,1) $: $ \sqrt{(2)^2 + 0} = 2 $
- To $ (2,2) $: $ \sqrt{0 + 1^2} = 1 $
Not consistent.

Wait — maybe it's not a semicircle.

Let’s try trapezoidal or triangular approximation?

But here's a better idea: observe that the graph is composed of straight lines and curves, but for simplicity, many such problems assume geometric shapes.

Wait — look at $ x=0 $ to $ x=4 $: from $ (0,1) $ to $ (2,2) $ to $ (4,0) $

So:
- From $ x=0 $ to $ x=2 $: line from $ (0,1) $ to $ (2,2) $
- From $ x=2 $ to $ x=4 $: line from $ (2,2) $ to $ (4,0) $

Wait — is it linear?

At $ x=0 $: $ f(0)=1 $
At $ x=2 $: $ f(2)=2 $
At $ x=4 $: $ f(4)=0 $

So yes! It’s two straight lines.

So from $ x=0 $ to $ x=2 $: increasing linearly from 1 to 2
From $ x=2 $ to $ x=4 $: decreasing linearly from 2 to 0

So the area from $ x=0 $ to $ x=4 $ is made of two triangles?

Wait — not exactly.

Let’s compute:

Area from $ x=0 $ to $ x=2 $:

- Triangle? From $ (0,1) $ to $ (2,2) $
- Base = 2, height change = 1
- But it's a trapezoid: average height $ \frac{1+2}{2} = 1.5 $, width = 2 → area = $ 1.5 \times 2 = 3 $

Or use integral: $ \int_0^2 f(x)\,dx $, where $ f(x) = \frac{1}{2}x + 1 $

Check: at $ x=0 $: $ f=1 $, at $ x=2 $: $ f= \frac{1}{2}(2)+1 = 2 $ → correct

So $ f(x) = \frac{1}{2}x + 1 $ on $ [0,2] $

Then:
$$
\int_0^2 \left(\frac{1}{2}x + 1\right) dx = \left[\frac{1}{4}x^2 + x\right]_0^2 = \frac{1}{4}(4) + 2 = 1 + 2 = 3
$$

Now $ x=2 $ to $ x=4 $: from $ (2,2) $ to $ (4,0) $

Slope: $ \frac{0 - 2}{4 - 2} = -1 $

So $ f(x) = -x + b $, plug in $ x=2 $: $ 2 = -2 + b \Rightarrow b=4 $

So $ f(x) = -x + 4 $ on $ [2,4] $

Integral:
$$
\int_2^4 (-x + 4)\,dx = \left[-\frac{1}{2}x^2 + 4x\right]_2^4 = \left(-\frac{1}{2}(16) + 16\right) - \left(-\frac{1}{2}(4) + 8\right) = (-8 + 16) - (-2 + 8) = 8 - 6 = 2
$$

So total from $ x=0 $ to $ x=4 $: $ 3 + 2 = 5 $

✔ So $ \int_0^4 f(x)\,dx = 5 $

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Interval $ [4,7] $: From $ x=4 $ to $ x=7 $

- From $ (4,0) $ to $ (5,1) $ to $ (6,2) $ to $ (7,2) $
- So:
- $ x=4 $ to $ x=5 $: line from $ (4,0) $ to $ (5,1) $: slope = 1 → $ f(x) = x - 4 $
- $ x=5 $ to $ x=6 $: line from $ (5,1) $ to $ (6,2) $: slope = 1 → $ f(x) = x - 4 $
Wait: at $ x=5 $: $ f=1 $, $ x=6 $: $ f=2 $ → $ f(x) = x - 4 $? At $ x=5 $: $ 5-4=1 $, yes; $ x=6 $: 2 → yes
- $ x=6 $ to $ x=7 $: horizontal line at $ y=2 $

So:
- $ x=4 $ to $ x=6 $: $ f(x) = x - 4 $
- $ x=6 $ to $ x=7 $: $ f(x) = 2 $

Compute:
$$
\int_4^6 (x - 4)\,dx = \left[\frac{1}{2}(x-4)^2\right]_4^6 = \frac{1}{2}(4) - 0 = 2
$$
$$
\int_6^7 2\,dx = 2 \cdot 1 = 2
$$
Total: $ 2 + 2 = 4 $

So $ \int_4^7 f(x)\,dx = 4 $

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Interval $ [7,9] $: From $ x=7 $ to $ x=9 $

- From $ (7,2) $ to $ (8,0) $ to $ (9,-2) $
- Line from $ (7,2) $ to $ (8,0) $: slope = $ -2 $
- $ f(x) = -2(x - 7) + 2 = -2x + 14 + 2 = -2x + 16 $
- From $ (8,0) $ to $ (9,-2) $: slope = $ -2 $
- $ f(x) = -2(x - 8) = -2x + 16 $

So same expression: $ f(x) = -2x + 16 $ on $ [7,9] $

Integrate:
$$
\int_7^9 (-2x + 16)\,dx = \left[-x^2 + 16x\right]_7^9 = [-(81) + 144] - [-(49) + 112] = (63) - (63) = 0
$$

Wait: $ -81 + 144 = 63 $, $ -49 + 112 = 63 $ → difference = 0

But visually: from $ x=7 $ to $ x=8 $: triangle above axis (area positive), then $ x=8 $ to $ x=9 $: triangle below axis (area negative). Since both have base 1, height 2, area $ \frac{1}{2}(1)(2) = 1 $ each → total area = $ 1 - 1 = 0 $

✔ So $ \int_7^9 f(x)\,dx = 0 $

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Interval $ [9,12] $: From $ x=9 $ to $ x=12 $

- From $ (9,-2) $ to $ (10,-1) $ to $ (11,0) $ to $ (12,0) $
- So:
- $ x=9 $ to $ x=10 $: line from $ (9,-2) $ to $ (10,-1) $: slope = 1 → $ f(x) = x - 11 $
- Check: $ x=9 $: $ 9 - 11 = -2 $, $ x=10 $: $ -1 $ → good
- $ x=10 $ to $ x=11 $: line from $ (10,-1) $ to $ (11,0) $: slope = 1 → $ f(x) = x - 11 $
- $ x=10 $: $ -1 $, $ x=11 $: 0 → yes
- $ x=11 $ to $ x=12 $: line from $ (11,0) $ to $ (12,0) $ → constant 0

So $ f(x) = x - 11 $ on $ [9,11] $, and 0 on $ [11,12] $

But $ f(x) = x - 11 $ is negative on $ [9,10] $, zero at $ x=11 $

So integrate:
$$
\int_9^{11} (x - 11)\,dx = \left[\frac{1}{2}(x - 11)^2\right]_9^{11} = 0 - \frac{1}{2}( -2)^2 = -\frac{1}{2}(4) = -2
$$
And $ \int_{11}^{12} 0\,dx = 0 $

So $ \int_9^{12} f(x)\,dx = -2 $

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Now sum all parts:

- $ \int_0^4 = 5 $
- $ \int_4^7 = 4 $
- $ \int_7^9 = 0 $
- $ \int_9^{12} = -2 $

So:
$$
\int_0^{12} f(x)\,dx = 5 + 4 + 0 + (-2) = \boxed{7}
$$

✔ Answer to #1: $ \boxed{7} $

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## ✔ Problem 2: $ \int_{12}^0 f(x)\,dx $

Use reversal of limits property:
$$
\int_{12}^0 f(x)\,dx = -\int_0^{12} f(x)\,dx = -7
$$

✔ Answer to #2: $ \boxed{-7} $

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## ✔ Problem 3: $ \int_7^6 f(x)\,dx $

Reversal of limits:
$$
\int_7^6 f(x)\,dx = -\int_6^7 f(x)\,dx
$$

From earlier, $ \int_6^7 f(x)\,dx = 2 $ (since $ f(x)=2 $ on $ [6,7] $)

So:
$$
\int_7^6 f(x)\,dx = -2
$$

✔ Answer to #3: $ \boxed{-2} $

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## ✔ Problem 4: $ \int_{12}^8 3f(x)\,dx $

First, reverse limits:
$$
\int_{12}^8 3f(x)\,dx = -\int_8^{12} 3f(x)\,dx = -3 \int_8^{12} f(x)\,dx
$$

Now compute $ \int_8^{12} f(x)\,dx $

Break into:
- $ [8,9] $: part of $ [7,9] $, already computed $ \int_7^9 = 0 $, so $ \int_8^9 = ? $

Wait — earlier we had $ \int_7^9 = 0 $, which includes:
- $ \int_7^8 $: triangle above axis, area $ \frac{1}{2}(1)(2) = 1 $
- $ \int_8^9 $: triangle below axis, area $ \frac{1}{2}(1)(2) = 1 $, but negative → $ -1 $
So $ \int_8^9 f(x)\,dx = -1 $

Then $ \int_9^{12} = -2 $ (from before)

So:
$$
\int_8^{12} f(x)\,dx = \int_8^9 + \int_9^{12} = (-1) + (-2) = -3
$$

Thus:
$$
\int_{12}^8 3f(x)\,dx = -3 \cdot (-3) = \boxed{9}
$$

✔ Answer to #4: $ \boxed{9} $

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## ✔ Problem 5: Given
- $ \int_{-2}^1 f(x)\,dx = 4 $
- $ \int_1^5 f(x)\,dx = -3 $
- $ \int_{-2}^1 g(x)\,dx = 8 $

Find:

a. $ \int_5^1 f(x)\,dx $

Reverse limits:
$$
\int_5^1 f(x)\,dx = -\int_1^5 f(x)\,dx = -(-3) = \boxed{3}
$$

b. $ \int_{-2}^5 f(x)\,dx $

Use addition property:
$$
\int_{-2}^5 f(x)\,dx = \int_{-2}^1 f(x)\,dx + \int_1^5 f(x)\,dx = 4 + (-3) = \boxed{1}
$$

c. $ \int_{-2}^1 [f(x) + 2g(x)]\,dx $

Use linearity:
$$
= \int_{-2}^1 f(x)\,dx + 2\int_{-2}^1 g(x)\,dx = 4 + 2(8) = 4 + 16 = \boxed{20}
$$

d. $ \int_0^1 f(x)\,dx $

We don’t have direct info. But we know $ \int_{-2}^1 f(x)\,dx = 4 $

Can't split without knowing $ \int_{-2}^0 f(x)\,dx $, unless we assume symmetry or something — but we don’t have data.

Wait — is there enough information?

No — we cannot determine $ \int_0^1 f(x)\,dx $ from given info.

But perhaps the graph is only defined for $ x \geq 0 $? But problem says $ \int_{-2}^1 $, so $ f $ is defined on $ [-2,1] $

But no graph for $ x < 0 $, so we cannot compute $ \int_0^1 f(x)\,dx $ unless we assume something.

But the problem gives $ \int_{-2}^1 f(x)\,dx = 4 $, but not $ \int_{-2}^0 $ or $ \int_0^1 $

So insufficient information?

Wait — unless the function is defined only on $ [0,12] $? But the integral from $ -2 $ to $ 1 $ suggests it’s defined there.

But no graph provided for $ x < 0 $, so likely we are expected to assume that $ f(x) $ is defined and we can use properties, but without additional info, we cannot find $ \int_0^1 f(x)\,dx $

But maybe the function is zero for $ x < 0 $? Unlikely.

Wait — perhaps the given integral $ \int_{-2}^1 f(x)\,dx = 4 $ is the only info, and we need to express in terms of others.

But unless we know $ \int_{-2}^0 f(x)\,dx $, we can’t get $ \int_0^1 $

So this cannot be determined from given info.

But let’s double-check: is there a typo? Or is $ f(x) $ assumed to be zero outside $ [0,12] $? But $ \int_{-2}^1 $ includes $ x<0 $, so probably not.

Wait — maybe the function is defined on $ [-2,1] $, but we don’t have graph.

So unless we are told $ f(x) = 0 $ for $ x < 0 $, we can’t compute.

But if we assume $ f(x) = 0 $ for $ x < 0 $, then $ \int_{-2}^0 f(x)\,dx = 0 $, so $ \int_0^1 f(x)\,dx = 4 $

But that’s an assumption.

Alternatively, perhaps the graph only shows $ x \geq 0 $, and $ f(x) $ is only defined for $ x \geq 0 $, so $ \int_{-2}^1 f(x)\,dx $ means $ \int_0^1 f(x)\,dx = 4 $? But that would require $ f(x)=0 $ for $ x<0 $

That’s a reasonable assumption in many contexts.

So assume $ f(x) = 0 $ for $ x < 0 $

Then $ \int_{-2}^1 f(x)\,dx = \int_0^1 f(x)\,dx = 4 $

So $ \int_0^1 f(x)\,dx = \boxed{4} $

✔ Answer to d: $ \boxed{4} $ *(assuming $ f(x) = 0 $ for $ x < 0 $)*

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e. $ \int_1^{-2} 3f(x)\,dx $

Reverse limits:
$$
= -\int_{-2}^1 3f(x)\,dx = -3 \int_{-2}^1 f(x)\,dx = -3(4) = \boxed{-12}
$$

f. $ \int_5^1 [f(x) - g(x)]\,dx $

Reverse limits:
$$
= -\int_1^5 [f(x) - g(x)]\,dx = -\left( \int_1^5 f(x)\,dx - \int_1^5 g(x)\,dx \right)
$$

We know $ \int_1^5 f(x)\,dx = -3 $

But we don’t know $ \int_1^5 g(x)\,dx $

We know $ \int_{-2}^1 g(x)\,dx = 8 $, but not $ \int_1^5 g(x)\,dx $

So cannot compute unless more info.

But wait — maybe we’re supposed to assume $ g(x) $ is defined, but no other info.

So insufficient information?

But perhaps the question expects us to leave it in terms of unknowns?

But that seems unlikely.

Wait — unless $ g(x) $ is related to $ f(x) $? But no.

So unless $ \int_1^5 g(x)\,dx $ is given, we can’t compute.

But perhaps the problem has a typo?

Alternatively, maybe $ g(x) $ is defined only on $ [-2,1] $? Then $ \int_1^5 g(x)\,dx = 0 $? But not stated.

So likely cannot be determined.

But perhaps we are meant to realize that $ \int_1^5 g(x)\,dx $ is unknown, so answer is undefined.

But let’s recheck: the only given is $ \int_{-2}^1 g(x)\,dx = 8 $, nothing else.

So $ \int_1^5 g(x)\,dx $ is unknown.

Thus, we cannot compute $ \int_5^1 [f(x) - g(x)]\,dx $ without more info.

But maybe the problem assumes $ g(x) = 0 $ for $ x > 1 $? Again, assumption.

If we assume $ g(x) = 0 $ for $ x > 1 $, then $ \int_1^5 g(x)\,dx = 0 $

Then:
$$
\int_5^1 [f(x) - g(x)]\,dx = -\left( \int_1^5 f(x)\,dx - \int_1^5 g(x)\,dx \right) = -(-3 - 0) = 3
$$

But this is assumption-based.

Without such assumption, cannot compute.

But given the context, perhaps the intended answer assumes $ g(x) = 0 $ for $ x > 1 $

So proceed with that.

Then:
$$
\int_5^1 [f(x) - g(x)]\,dx = -\left( -3 - 0 \right) = \boxed{3}
$$

But strictly speaking, insufficient information.

However, in many textbook problems, if a function is only defined on a certain interval, it’s assumed to be zero elsewhere.

So assuming $ g(x) = 0 $ for $ x > 1 $, then $ \int_1^5 g(x)\,dx = 0 $

✔ Answer to f: $ \boxed{3} $ *(with assumption)*

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## ✔ Summary of Answers:

1. $ \int_0^{12} f(x)\,dx = \boxed{7} $
2. $ \int_{12}^0 f(x)\,dx = \boxed{-7} $
3. $ \int_7^6 f(x)\,dx = \boxed{-2} $
4. $ \int_{12}^8 3f(x)\,dx = \boxed{9} $
5.
a. $ \int_5^1 f(x)\,dx = \boxed{3} $
b. $ \int_{-2}^5 f(x)\,dx = \boxed{1} $
c. $ \int_{-2}^1 [f(x) + 2g(x)]\,dx = \boxed{20} $
d. $ \int_0^1 f(x)\,dx = \boxed{4} $ *(assuming $ f(x) = 0 $ for $ x < 0 $)*
e. $ \int_1^{-2} 3f(x)\,dx = \boxed{-12} $
f. $ \int_5^1 [f(x) - g(x)]\,dx = \boxed{3} $ *(assuming $ g(x) = 0 $ for $ x > 1 $)*

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## 📌 Notes on Properties (Fill in the blanks):

| Property | Formula |
|--------|--------|
| Equivalent Limits | $ \int_a^a f(x)\,dx = \boxed{0} $ |
| Reversal of Limits | $ \int_a^b f(x)\,dx = \boxed{-\int_b^a f(x)\,dx} $ |
| Multiply by constant | $ \int_a^b k f(x)\,dx = \boxed{k \int_a^b f(x)\,dx} $ |
| Adjacent Intervals | $ \int_a^c f(x)\,dx + \int_c^b f(x)\,dx = \boxed{\int_a^b f(x)\,dx} $ |
| Addition | $ \int_a^b [f(x) + g(x)]\,dx = \boxed{\int_a^b f(x)\,dx + \int_a^b g(x)\,dx} $ |
| Subtraction | $ \int_a^b [f(x) - g(x)]\,dx = \boxed{\int_a^b f(x)\,dx - \int_a^b g(x)\,dx} $ |

These are standard properties of definite integrals.

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Let me know if you want a visual explanation or clarification on any part!