How do you find the average value of the function for #f(x)=2xsqrt(1+x^2), -3

Question

How do you find the average value of the function for #f(x)=2xsqrt(1+x^2), -3<=x<=3#?

William Abdalla · Accepted Answer

0

To find the average, we take the integral and divide through the length of the interval. F(x) = #int f(x) dx # = #int sqrt(1+x²) d(1+x²) # = #(2/3) (1+x²)^(3/2)# Now evaluate between -3 and 3 : F(3) - F(-3) = 0 So 0/6 = 0.

Samantha Adkison · Answer

#0# The average value of a function over an interval is equal to the definite integral of that interval divided by the length of the interval. If we wanted to find the average value of #f(x)# on the interval #[a,b]#, we can express it in general terms like this: #(int_a^b f(x)\ dx)/(b-a)# If we plug in our function, we get: #(int_-3^3 2xsqrt(1+x^2)\ dx)/(3-(-3))# Let's first start by computing the anti-derivative of the function. We can quite quickly see that we have the derivative of #1+x^2#, #2x#, on the outside of the square root. This is a tell-tale sign that we can use u-substitution with #u=1+x^2#. We knew that the derivative was #2x#, so we can just divide through by #2x#: #int\ 2xsqrt(1+x^2)\ dx=int\ (cancel(2x)sqrt(1+x^2))/cancel(2x)\ du=int\ sqrtu\ du# By knowing #sqrt(u)=u^(1/2)#, we can use the power rule: #int\ sqrtu\ du=int\ u^(1/2)\ du=u^(3/2)/(3/2)=2/3u^(3/2)=2/3(1+x^2)^(3/2)# Now we can evaluate the definite integral: #int_-3^3 2xsqrt(1+x^2)\ dx=[2/3(1+x^2)^(3/2)]_-3^3# #=2/3 (1+3^2)^(3/2)-2/3(1+(-3)^2)^(3/2)=0# Now we divide by #3-(-3)=6#, which gives: #0/6=0# So, the average value of the function is #0# on the interval #[-3,3]#. In fact, since the function is mirrored upside down on the other side of the x-axis you can say that for any real number #a#, the average value on the interval #[-a,a]# will be equal to #0#.

Emilia Aguilar · Answer

undefined To find the average value of the function $ f(x) = 2x \sqrt{1+x^2} $ over the interval $[-3, 3]$, we use the formula for average value of a function over a closed interval:

$$ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx $$

In this case, $ a = -3 $ and $ b = 3 $.

$$ \text{Average value} = \frac{1}{3 - (-3)} \int_{-3}^{3} 2x \sqrt{1+x^2} \, dx $$

Now, we need to compute the integral:

$$ \int_{-3}^{3} 2x \sqrt{1+x^2} \, dx $$

This integral can be evaluated using a trigonometric substitution. Let $ u = 1 + x^2 $. Then $ du = 2x \, dx $. When $ x = -3 $, $ u = 10 $, and when $ x = 3 $, $ u = 10 $. Therefore, the integral becomes:

$$ \int_{10}^{10} \sqrt{u} \, du $$

$$ = \left[ \frac{2}{3} u^{3/2} \right]_{10}^{10} $$

$$ = \frac{2}{3}(10)^{3/2} - \frac{2}{3}(10)^{3/2} $$

$$ = 0 $$

Since the integral evaluates to 0, the average value of the function over the interval $[-3, 3]$ is also 0.

Emily Ammerman · Answer

undefined To find the average value of the function $ f(x) = 2x\sqrt{1 + x^2} $ over the interval $[-3, 3]$, you can use the formula for the average value of a function over an interval $[a, b]$:

$$ \text{Average value} = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx $$

Substituting $ f(x) = 2x\sqrt{1 + x^2} $, $ a = -3 $, and $ b = 3 $ into this formula:

$$ \text{Average value} = \frac{1}{3 - (-3)} \int_{-3}^{3} 2x\sqrt{1 + x^2} \, dx $$

$$ \text{Average value} = \frac{1}{6} \int_{-3}^{3} 2x\sqrt{1 + x^2} \, dx $$

This integral can be challenging to solve directly, but we can simplify it by using a substitution. Let $ u = 1 + x^2 $, then $ du = 2x \, dx $. When $ x = -3 $, $ u = 1 + (-3)^2 = 10 $, and when $ x = 3 $, $ u = 1 + 3^2 = 10 $. So, the integral becomes:

$$ \text{Average value} = \frac{1}{6} \int_{10}^{10} \sqrt{u} \, du $$

$$ \text{Average value} = \frac{1}{6} \left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{10}^{10} $$

$$ \text{Average value} = \frac{1}{6} \left( \frac{2}{3} \cdot 10^{\frac{3}{2}} - \frac{2}{3} \cdot 10^{\frac{3}{2}} \right) $$

$$ \text{Average value} = \frac{1}{6} \cdot 0 $$

$$ \text{Average value} = 0 $$

Therefore, the average value of the function $ f(x) = 2x\sqrt{1 + x^2} $ over the interval $[-3, 3]$ is $ 0 $.

How do you find the average value of the function for #f(x)=2xsqrt(1+x^2), -3&lt;=x&lt;=3#?

How do you find the average value of the function for #f(x)=2xsqrt(1+x^2), -3<=x<=3#?