Basic definite integral question

In summary, the person is asking for confirmation on their approach to finding the average value of h(x) using integrals. They suggest adding or subtracting definite integrals to get the desired range and have found the average value to be 17/8. They ask for confirmation on their method and provide a possible solution using integrals.
  • #1
TheFallen018
52
0
Hey, I've got this problem I've been doing, but I'm not sure if my approach is right. My textbook has pretty much less than a paragraph on this sort of stuff.

View attachment 8444

My thinking was that since an integral is a sum, in order to get the range from 0 to 8, we should just be able to add or subtract the definite integrals as required. eg. $\int_{0}^{5}f(\Theta )d\Theta + \int_{5}^{8}f(\Theta)d\Theta = \int_{0}^{8}f(\Theta)d\Theta = 3+8=11$

So, following this line of thinking, I found $\int_{0}^{8}h(x)dx$ to be $2*11-5=17$. Then, by dividing by 8, we should get the average value of $h(x)$ to be ($\frac{17}{8}$).

It feels like I'm missing something though. Does this look right? Thanks
 

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  • #2
Yes, you would need to do this:
\begin{align*}
\langle h\rangle&=\frac18\int_0^8h(x)\,dx \\
&=\frac18\int_0^8(2f(x)-g(x))\,dx \\
&=\frac18\left[2\int_0^8f(x)\,dx-\int_0^8g(x)\,dx\right] \\
&=\frac18\left[2\left(\int_0^5f(x)\,dx+\int_5^8f(x)\,dx\right)-\left(\int_0^{10}g(x)\,dx-\int_8^{10}g(x)\,dx\right)\right].
\end{align*}
Then you can plug in what you know. Note: the $\langle h\rangle$ notation means "the average value of $h$."
 

Related to Basic definite integral question

1. What is a definite integral?

A definite integral is a mathematical concept that represents the area under a curve between two points on a graph. It is used to find the total value of a function over a specific interval.

2. How do you solve a basic definite integral?

To solve a basic definite integral, you first need to identify the limits of integration, which are the two points that define the interval. Then, you need to integrate the function using the rules of integration. Finally, you evaluate the integral using the limits of integration to find the total value of the function over the interval.

3. What is the difference between a definite integral and an indefinite integral?

The main difference between a definite integral and an indefinite integral is that the former has specific limits of integration, while the latter does not. A definite integral gives a numerical value, while an indefinite integral gives a function.

4. Why is the definite integral important in calculus?

The definite integral is important in calculus because it allows us to find the total value of a function over a specific interval. This is useful in many real-world applications, such as calculating areas, volumes, and averages.

5. Can the definite integral be negative?

Yes, the definite integral can be negative if the function is below the x-axis in the given interval. This represents a negative area under the curve, which can still be useful in certain applications.

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