Answer :
Certainly! Let's evaluate the definite integral [tex]\(\int_0^{\frac{x}{2}} \sin(t) \, dt\)[/tex], where the variable of integration is [tex]\( t \)[/tex]. Here is a detailed step-by-step solution:
1. Identify the Integrand:
The integrand in this problem is [tex]\(\sin(t)\)[/tex].
2. Find the Indefinite Integral:
To integrate [tex]\(\sin(t)\)[/tex] with respect to [tex]\( t \)[/tex], we find the antiderivative of [tex]\(\sin(t)\)[/tex]:
[tex]\[ \int \sin(t) \, dt = -\cos(t) + C \][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration, which we will consider when evaluating the definite integral.
3. Set Up the Definite Integral:
We need to evaluate the definite integral from the lower bound [tex]\( t = 0 \)[/tex] to the upper bound [tex]\( t = \frac{x}{2} \)[/tex]:
[tex]\[ \int_0^{\frac{x}{2}} \sin(t) \, dt \][/tex]
4. Evaluate the Antiderivative at the Bounds:
Using the fundamental theorem of calculus, we substitute the bounds into the antiderivative:
[tex]\[ \left[ -\cos(t) \right]_0^{\frac{x}{2}} \][/tex]
This means we compute:
[tex]\[ -\cos\left(\frac{x}{2}\right) - \left( -\cos(0) \right) \][/tex]
5. Simplify the Expression:
We know that [tex]\(\cos(0) = 1\)[/tex], so the expression simplifies to:
[tex]\[ -\cos\left(\frac{x}{2}\right) + 1 \][/tex]
6. Rewrite for Clarity:
Therefore, the definite integral evaluates to:
[tex]\[ 1 - \cos\left(\frac{x}{2}\right) \][/tex]
So, the final result of the integral [tex]\(\int_0^{\frac{x}{2}} \sin(t) \, dt\)[/tex] is:
[tex]\[ 1 - \cos\left(\frac{x}{2}\right) \][/tex]
1. Identify the Integrand:
The integrand in this problem is [tex]\(\sin(t)\)[/tex].
2. Find the Indefinite Integral:
To integrate [tex]\(\sin(t)\)[/tex] with respect to [tex]\( t \)[/tex], we find the antiderivative of [tex]\(\sin(t)\)[/tex]:
[tex]\[ \int \sin(t) \, dt = -\cos(t) + C \][/tex]
Here, [tex]\(C\)[/tex] is the constant of integration, which we will consider when evaluating the definite integral.
3. Set Up the Definite Integral:
We need to evaluate the definite integral from the lower bound [tex]\( t = 0 \)[/tex] to the upper bound [tex]\( t = \frac{x}{2} \)[/tex]:
[tex]\[ \int_0^{\frac{x}{2}} \sin(t) \, dt \][/tex]
4. Evaluate the Antiderivative at the Bounds:
Using the fundamental theorem of calculus, we substitute the bounds into the antiderivative:
[tex]\[ \left[ -\cos(t) \right]_0^{\frac{x}{2}} \][/tex]
This means we compute:
[tex]\[ -\cos\left(\frac{x}{2}\right) - \left( -\cos(0) \right) \][/tex]
5. Simplify the Expression:
We know that [tex]\(\cos(0) = 1\)[/tex], so the expression simplifies to:
[tex]\[ -\cos\left(\frac{x}{2}\right) + 1 \][/tex]
6. Rewrite for Clarity:
Therefore, the definite integral evaluates to:
[tex]\[ 1 - \cos\left(\frac{x}{2}\right) \][/tex]
So, the final result of the integral [tex]\(\int_0^{\frac{x}{2}} \sin(t) \, dt\)[/tex] is:
[tex]\[ 1 - \cos\left(\frac{x}{2}\right) \][/tex]