What is #f(x) = int x+3xsqrt(x^2+1) dx# if #f(2) = 7 #?

Question

What is #f(x) = int x+3xsqrt(x^2+1)   dx# if #f(2) = 7   #?

Emma Aldridge · Accepted Answer

#\implies f(x) = x^2/2+(x^2+1)^{3/2} + 5(1- sqrt5)# #f(x) = int x+3xsqrt(x^2+1) dx# noting the pattern #D ( alpha (x^2+1)^{3/2} ) = alpha 3/2 (x^2+1)^{1/2} 2x = 3 alpha x (x^2+1)^{1/2} # so here #alpha = 1# #\implies f(x) = x^2/2+(x^2+1)^{3/2} +C# #7= 2+(5)^{3/2} +C implies C = 5(1- sqrt5)# #\implies f(x) = x^2/2+(x^2+1)^{3/2} + 5(1- sqrt5)#

Ezra Adams · Answer

undefined The integral of $x + 3x\sqrt{x^2 + 1}$ with respect to $x$ is $ \frac{x^2}{2} + \frac{3}{5}(x^2 + 1)^{\frac{3}{2}} + C$.

Given that $f(2) = 7$, we can solve for $C$:

$$7 = \frac{2^2}{2} + \frac{3}{5}(2^2 + 1)^{\frac{3}{2}} + C$$

$$7 = 2 + \frac{3}{5}(5)^{\frac{3}{2}} + C$$

$$7 = 2 + \frac{3}{5}(5\sqrt{5}) + C$$

$$7 = 2 + 3\sqrt{5} + C$$

$$C = 7 - 2 - 3\sqrt{5}$$

$$C = 5 - 3\sqrt{5}$$

Therefore, the function $f(x)$ is:

$$f(x) = \frac{x^2}{2} + \frac{3}{5}(x^2 + 1)^{\frac{3}{2}} + 5 - 3\sqrt{5}$$

Samantha Adkison · Answer

undefined The function $ f(x) = \int (x + 3x\sqrt{x^2 + 1}) \, dx $ represents the integral of the expression $ (x + 3x\sqrt{x^2 + 1}) $ with respect to $ x $. Given that $ f(2) = 7 $, we need to find the value of the integral when $ x = 2 $.

First, we need to find the antiderivative of the expression inside the integral:

$$ \int (x + 3x\sqrt{x^2 + 1}) \, dx $$

$$ = \int x \, dx + \int 3x\sqrt{x^2 + 1} \, dx $$

$$ = \frac{x^2}{2} + \int 3x\sqrt{x^2 + 1} \, dx $$

To find $ \int 3x\sqrt{x^2 + 1} \, dx $, we can use substitution. Let $ u = x^2 + 1 $, then $ du = 2x \, dx $, or $ \frac{du}{2} = x \, dx $.

$$ \int 3x\sqrt{x^2 + 1} \, dx = \int 3\sqrt{u} \, \frac{du}{2} $$

$$ = \frac{3}{2} \int \sqrt{u} \, du $$

$$ = \frac{3}{2} \cdot \frac{2}{3} u^{3/2} + C $$

$$ = u^{3/2} + C $$

$$ = (x^2 + 1)^{3/2} + C $$

Therefore, the antiderivative of $ f(x) $ is $ F(x) = \frac{x^2}{2} + (x^2 + 1)^{3/2} + C $, where $ C $ is the constant of integration.

Now, we can find the value of $ F(2) $:

$$ F(2) = \frac{2^2}{2} + (2^2 + 1)^{3/2} + C $$

$$ = 2 + (5)^{3/2} + C $$

Given that $ f(2) = 7 $, we can solve for $ C $:

$$ 7 = 2 + (5)^{3/2} + C $$

$$ C = 7 - 2 - (5)^{3/2} $$

$$ C = 7 - 2 - 5\sqrt{5} $$

$$ C = 5 - 5\sqrt{5} $$

Therefore, the value of the integral $ f(x) $ when $ x = 2 $ is $ 2 + (5)^{3/2} + 5 - 5\sqrt{5} $.