Math / Calculus

QuestionQuestion 8
Use the identity $\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x$ to compute the integral. $\int_{-\pi}^{\pi} \frac{9 \sin (t)}{25+t^{2}} d t=$ $\square$ Hint: $\sin (-t)=-\sin (t)$ . Submit Question

Studdy Solution

STEP 1

What is this asking? We need to calculate a definite integral of a trigonometric function over a symmetric interval, using a given identity and a hint. Watch out! Don't forget about that helpful hint!
It's there for a reason!
Also, remember that the integral of an odd function over a symmetric interval is zero.

STEP 2

1. Analyze the function
2. Apply the identity
3. Compute the integral

STEP 3

Let's define our function $f(t) = \frac{9 \sin(t)}{25 + t^2}$ .
We're going to figure out if this function is odd or even!
This is super important because it can make our calculation much easier.

STEP 4

Remember, a function is odd if $f(-t) = -f(t)$ .
Let's test it out!

STEP 5

We have $f(-t) = \frac{9 \sin(-t)}{25 + (-t)^2} = \frac{9 \sin(-t)}{25 + t^2}$ .
Now, the hint tells us that $\sin(-t) = -\sin(t)$ , so we can substitute that in: $f(-t) = \frac{-9 \sin(t)}{25 + t^2}$ .
Look closely – that's just $-f(t)$ !

STEP 6

So, $f(t)$ is an odd function!
This is awesome because the integral of an odd function over a symmetric interval (like $[-\pi, \pi]$ ) is always zero!

STEP 7

The problem gives us this handy identity: $\int_{-A}^{A} f(x) dx = \int_{-A}^{0} f(x) dx + \int_{0}^{A} f(x) dx$ .
In our case, $A = \pi$ and $f(x)$ is replaced by $f(t)$ .
Even though we already know the answer from analyzing the function, let's apply the identity to see how it works with the given hint.

STEP 8

So, we have $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = \int_{-\pi}^{0} \frac{9 \sin(t)}{25 + t^2} dt + \int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt$ .

STEP 9

Since $f(t)$ is an odd function, we know that $\int_{-\pi}^{0} f(t) dt = -\int_{0}^{\pi} f(t) dt$ .

STEP 10

Substituting this into our equation from the previous step, we get: $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = -\int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt + \int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt$ .

STEP 11

Look! We're adding a value to its negative, which means they add to zero!
So, $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = 0$ .

STEP 12

The value of the definite integral is 0.

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QuestionQuestion 8
Use the identity $\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x$ to compute the integral. $\int_{-\pi}^{\pi} \frac{9 \sin (t)}{25+t^{2}} d t=$ $\square$ Hint: $\sin (-t)=-\sin (t)$ . Submit Question

Studdy Solution

STEP 1

STEP 2

1. Analyze the function
2. Apply the identity
3. Compute the integral

STEP 3

Let's define our function $f(t) = \frac{9 \sin(t)}{25 + t^2}$ .
We're going to figure out if this function is odd or even!
This is super important because it can make our calculation much easier.

STEP 4

Remember, a function is odd if $f(-t) = -f(t)$ .
Let's test it out!

STEP 5

We have $f(-t) = \frac{9 \sin(-t)}{25 + (-t)^2} = \frac{9 \sin(-t)}{25 + t^2}$ .
Now, the hint tells us that $\sin(-t) = -\sin(t)$ , so we can substitute that in: $f(-t) = \frac{-9 \sin(t)}{25 + t^2}$ .
Look closely – that's just $-f(t)$ !

STEP 6

So, $f(t)$ is an odd function!
This is awesome because the integral of an odd function over a symmetric interval (like $[-\pi, \pi]$ ) is always zero!

STEP 7

STEP 8

So, we have $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = \int_{-\pi}^{0} \frac{9 \sin(t)}{25 + t^2} dt + \int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt$ .

STEP 9

Since $f(t)$ is an odd function, we know that $\int_{-\pi}^{0} f(t) dt = -\int_{0}^{\pi} f(t) dt$ .

STEP 10

Substituting this into our equation from the previous step, we get: $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = -\int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt + \int_{0}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt$ .

STEP 11

Look! We're adding a value to its negative, which means they add to zero!
So, $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = 0$ .

STEP 12

The value of the definite integral is 0.

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Studdy Solution

STEP 1

STEP 2

1. Analyze the function2. Apply the identity3. Compute the integral

STEP 3

Let's **define** our function f(t)=9sin⁡(t)25+t2f(t) = \frac{9 \sin(t)}{25 + t^2}f(t)=25+t29sin(t)​.We're going to figure out if this function is odd or even!This is super important because it can make our calculation much easier.

STEP 4

Remember, a function is *odd* if f(−t)=−f(t)f(-t) = -f(t)f(−t)=−f(t).Let's test it out!

STEP 5

STEP 6

So, f(t)f(t)f(t) is an **odd function**!This is awesome because the integral of an odd function over a symmetric interval (like [−π,π][-\pi, \pi][−π,π]) is always **zero**!

STEP 7

STEP 8

STEP 9

Since f(t)f(t)f(t) is an odd function, we know that ∫−π0f(t)dt=−∫0πf(t)dt\int_{-\pi}^{0} f(t) dt = -\int_{0}^{\pi} f(t) dt∫−π0​f(t)dt=−∫0π​f(t)dt.

STEP 10

STEP 11

Look! We're adding a value to its negative, which means they add to zero!So, ∫−ππ9sin⁡(t)25+t2dt=0\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = 0∫−ππ​25+t29sin(t)​dt=0.

STEP 12

The value of the definite integral is **0**.

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Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Analyze the function
2. Apply the identity
3. Compute the integral

Let's define our function $f(t) = \frac{9 \sin(t)}{25 + t^2}$ .
We're going to figure out if this function is odd or even!
This is super important because it can make our calculation much easier.

Remember, a function is odd if $f(-t) = -f(t)$ .
Let's test it out!

So, $f(t)$ is an odd function!
This is awesome because the integral of an odd function over a symmetric interval (like $[-\pi, \pi]$ ) is always zero!

Since $f(t)$ is an odd function, we know that $\int_{-\pi}^{0} f(t) dt = -\int_{0}^{\pi} f(t) dt$ .

Look! We're adding a value to its negative, which means they add to zero!
So, $\int_{-\pi}^{\pi} \frac{9 \sin(t)}{25 + t^2} dt = 0$ .

The value of the definite integral is 0.