Evaluate The Integral: ∫ 1 4 T − 3 / 2 D T \int_1^4 T^{-3 / 2} \, Dt ∫ 1 4 ​ T − 3/2 D T Options:A. -1

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Introduction

In this article, we will evaluate the given integral $\int_1^4 t^{-3 / 2} \, dt$. This involves finding the antiderivative of the function $t^{-3 / 2}$ and then applying the fundamental theorem of calculus to determine the definite integral. We will use the power rule of integration and other techniques to find the antiderivative.

The Power Rule of Integration

The power rule of integration states that if $f(x) = x^n$, then $\int f(x) \, dx = \frac{x^{n+1}}{n+1} + C$. This rule can be applied to integrate functions of the form $t^n$, where $n$ is a constant.

Evaluating the Integral

To evaluate the integral $\int_1^4 t^{-3 / 2} \, dt$, we can use the power rule of integration. We have $t^{-3 / 2} = t^{-3} \cdot t^{-1/2}$. Using the power rule, we can integrate each of these terms separately.

Step 1: Integrate $t^{-3}$

Using the power rule, we have $\int t^{-3} \, dt = \frac{t^{-2}}{-2} + C = -\frac{1}{2t^2} + C$.

Step 2: Integrate $t^{-1/2}$

Using the power rule, we have $\int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} + C = 2t^{1/2} + C$.

Step 3: Combine the Results

We can now combine the results from steps 1 and 2 to find the antiderivative of $t^{-3 / 2}$. We have $\int t^{-3 / 2} \, dt = -\frac{1}{2t^2} + 2t^{1/2} + C$.

Step 4: Apply the Fundamental Theorem of Calculus

To find the definite integral, we can apply the fundamental theorem of calculus. We have $\int_1^4 t^{-3 / 2} \, dt = \left[-\frac{1}{2t^2} + 2t^{1/2}\right]_1^4$.

Step 5: Evaluate the Expression

We can now evaluate the expression by substituting the limits of integration. We have $\left[-\frac{1}{2t^2} + 2t^{1/2}\right]_1^4 = \left(-\frac{1}{2(4)^2} + 2(4)^{1/2}\right) - \left(-\frac{1}{2(1)^2} + 2(1)^{1/2}\right)$.

Step 6: Simplify the Expression

We can now simplify the expression by evaluating the terms. We have $\left(-\frac{1}{2(4)^2} + 2(4)^{1/2}\right) - \left(-\frac{1}{2(1)^2} + 2(1)^{1/2}\right) = \left(-\frac{1}{32} + 4\right) - \left(-\frac{1}{2} + 2\right)$.

Step 7: Combine the Terms

We can now combine the terms to find the final answer. We have $\left(-\frac{1}{32} + 4\right) - \left(-\frac{1}{2} + 2\right) = -\frac{1}{32} + 4 + \frac{1}{2} - 2$.

Step 8: Simplify the Expression

We can now simplify the expression by combining the terms. We have $-\frac{1}{32} + 4 + \frac{1}{2} - 2 = -\frac{1}{32} + \frac{1}{2} + 2$.

Step 9: Combine the Terms

We can now combine the terms to find the final answer. We have $-\frac{1}{32} + \frac{1}{2} + 2 = -\frac{1}{32} + \frac{16}{32} + \frac{64}{32}$.

Step 10: Simplify the Expression

We can now simplify the expression by combining the terms. We have $-\frac{1}{32} + \frac{16}{32} + \frac{64}{32} = \frac{79}{32}$.

The final answer is: $\boxed{\frac{79}{32}}$

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Introduction

In our previous article, we evaluated the integral $\int_1^4 t^{-3 / 2} \, dt$ using the power rule of integration and the fundamental theorem of calculus. In this article, we will answer some common questions related to this integral.

Q: What is the antiderivative of $t^{-3 / 2}$?

A: The antiderivative of $t^{-3 / 2}$ is $-\frac{1}{2t^2} + 2t^{1/2} + C$.

Q: How do I apply the fundamental theorem of calculus to evaluate the definite integral?

A: To apply the fundamental theorem of calculus, you need to substitute the limits of integration into the antiderivative and then subtract the value of the antiderivative at the lower limit from the value of the antiderivative at the upper limit.

Q: What is the value of the definite integral $\int_1^4 t^{-3 / 2} \, dt$?

A: The value of the definite integral $\int_1^4 t^{-3 / 2} \, dt$ is $\frac{79}{32}$.

Q: Can I use the power rule of integration to integrate $t^{-3 / 2}$?

A: Yes, you can use the power rule of integration to integrate $t^{-3 / 2}$. The power rule states that if $f(x) = x^n$, then $\int f(x) \, dx = \frac{x^{n+1}}{n+1} + C$.

Q: What is the significance of the constant $C$ in the antiderivative?

A: The constant $C$ in the antiderivative is an arbitrary constant that represents the family of antiderivatives of the function. It is not a part of the definite integral.

Q: Can I use the fundamental theorem of calculus to evaluate the definite integral of any function?

A: Yes, you can use the fundamental theorem of calculus to evaluate the definite integral of any function. However, you need to find the antiderivative of the function first.

Q: What is the relationship between the definite integral and the antiderivative?

A: The definite integral is the difference between the value of the antiderivative at the upper limit and the value of the antiderivative at the lower limit.

Q: Can I use the power rule of integration to integrate any function?

A: Yes, you can use the power rule of integration to integrate any function of the form $f(x) = x^n$, where $n$ is a constant.

Q: What is the significance of the exponent $-3/2$ in the integral?

A: The exponent $-3/2$ in the integral represents the power to which the variable $t$ is raised. It is an important part of the integral and affects the value of the definite integral.

Q: Can I use the fundamental theorem of calculus to evaluate the definite integral of a function that is not continuous?

A: No, you cannot use the fundamental theorem of calculus to evaluate the definite integral of a function that is not continuous. The fundamental theorem of calculus requires the function to be continuous on the interval of integration.

Q: What is the relationship between the definite integral and the area under the curve?

A: The definite integral represents the area under the curve of the function on the interval of integration.

Q: Can I use the power rule of integration to integrate a function that is not of the form $f(x) = x^n$?

A: No, you cannot use the power rule of integration to integrate a function that is not of the form $f(x) = x^n$. The power rule is only applicable to functions of this form.

Q: What is the significance of the limits of integration in the definite integral?

A: The limits of integration in the definite integral represent the interval on which the function is being integrated. They are an important part of the integral and affect the value of the definite integral.

Q: Can I use the fundamental theorem of calculus to evaluate the definite integral of a function that is not defined on the interval of integration?

A: No, you cannot use the fundamental theorem of calculus to evaluate the definite integral of a function that is not defined on the interval of integration. The fundamental theorem of calculus requires the function to be defined on the interval of integration.