# Fundamental Theorem of Caluculus

• January 22nd 2013, 01:34 PM
irock3337
Fundamental Theorem of Caluculus
x
G(x)=∫ t((3t^2+4)^1/2)dt
4

Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)
• January 22nd 2013, 01:43 PM
Plato
Re: Fundamental Theorem of Caluculus
Quote:

Originally Posted by irock3337
x
G(x)=∫ t((3t^2+4)^1/2)dt
4
Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)

What is the derivative of $\tfrac{1}{9}(3t^2+4)^{3/2}~?$
• January 22nd 2013, 01:51 PM
irock3337
Re: Fundamental Theorem of Caluculus
t(3t^2+4)^2 or is it t(3t^2+4)^1/2
• January 22nd 2013, 02:00 PM
Plato
Re: Fundamental Theorem of Caluculus
Quote:

Originally Posted by irock3337
t(3t^2+4)^2 or is it t(3t^2+4)^1/2

It is $t(3t^2+4)^{1/2}$
• January 22nd 2013, 02:10 PM
irock3337
Re: Fundamental Theorem of Caluculus
So then would I just plug x and 4 into the equation and subtract them from one another to get the answer?
• January 22nd 2013, 04:02 PM
Prove It
Re: Fundamental Theorem of Caluculus
Quote:

Originally Posted by irock3337
x
G(x)=∫ t((3t^2+4)^1/2)dt
4

Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)

If you were going to integrate this directly, first rewrite this integral as \displaystyle \begin{align*} \int_4^x{t\left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} = \frac{1}{6}\int_4^x{6t\left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} \end{align*} and then use a substitution \displaystyle \begin{align*} u = 3t^2 + 4 \implies du = 6t\,dt \end{align*}, noting that when \displaystyle \begin{align*} t = 4, u = 52 \end{align*} and when \displaystyle \begin{align*} t = x, u = 3x^2 + 4 \end{align*}, and the integral becomes

\displaystyle \begin{align*} \frac{1}{6}\int_4^x{6t \left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} &= \frac{1}{6}\int_{52}^{3x^2 + 4}{u^{\frac{1}{2}}\,du} \\ &= \frac{1}{6}\left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{52}^{3x^2 + 4} \\ &= \frac{1}{9} \left[ \left( 3x^2 + 4 \right)^{\frac{3}{2}} - 52^{\frac{3}{2}} \right] \end{align*}