# *Use substitution to evaluate the following integral, showing all of your work: \int...

## Question:

• Use substitution to evaluate the following integral, showing all of your work:

{eq}\int \frac{4}{(1+2x)^{2}}dx {/eq}

• Let

{eq}\oint_{3}^{x} \sqrt{3t^{2}+2}dt {/eq} and find {eq}F(3), {F}'(3), \ and \ {F}''(3). {/eq}

## U Substitution:

One integration technique is known as the reverse chain rule, or more commonly known as u-substitution. This method involves the substitution of the integrand to a dummy variable in order to make the process of integration easier. The substitution is reverted after the integration process.

For the first problem, we can allow a substitution, {eq}\displaystyle u = 1+2x {/eq} which makes {eq}\displaystyle du = 2dx {/eq} or {eq}\displaystyle \frac{du}{2} = dx {/eq}. We proceed with the integration over the transformed variable and then revert the substitution afterwards.

{eq}\begin{align} \displaystyle \int \frac{4}{(1+2x)^2}dx &= 2\int \frac{1}{u^2}du\\ &= 2\int u^{-2}du\\ &= -2u^{-1} +C\\ &= - \frac{2}{1+2x} +C \end{align} {/eq}

For the next problem, we evaluate the function separately. We find {eq}\displaystyle F(3) {/eq}.

{eq}\begin{align} \displaystyle F(3) &= \oint_3^3 \sqrt{3t^2+2}dt \end{align} {/eq}

Since we have the same upper and lower limits for integration, {eq}\displaystyle F(3) = \oint_3^3 \sqrt{3t^2+2}dt = 0 {/eq}.

To evaluate {eq}\displaystyle F'(3) {/eq} we must assign {eq}\displaystyle g(t) = \sqrt{3t^2+2} {/eq} with the antiderivative function, {eq}\displaystyle G(t) {/eq} such that {eq}\displaystyle G'(t) = G(t) {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle F(x) &= \int_x^3 f(t)dt\\ &= G(3)- G(x)\\ F'(x) &= \frac{d}{dx} (G(3) - G(x))\\ &= 0 - G'(x)\\ &= -g(x)\\ &= -\sqrt{3x^2+2}\\ F'(3) &= -\sqrt{3(3)^2 +2}\\ &= -\sqrt{29} \end{align} {/eq}

Finally, we find {eq}\displaystyle F''(3) {/eq} by first differentiating {eq}\displaystyle F'(x) {/eq} and then evaluating it at {eq}\displaystyle x=3 {/eq}. We use the chain rule, {eq}\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle F'(x)& =-\sqrt{3x^2+2}\\ &= -(3x^2+2)^ \frac{1}{2}\\ F''(x) &=-\frac{d (3x^2+2)^\frac{1}{2}}{d(3x^2+2)} \frac{d(3x^2+2)}{dx}\\ &= -\frac{1(3x^2+2)^{- \frac{1}{2}}}{2} \cdot (6x)\\ &= -\frac{3x}{\sqrt{3x^2+2}}\\ F''(3) &= -\frac{3(3)}{\sqrt{3(3)^2+2}}\\ &=- \frac{9}{\sqrt{29}} \end{align} {/eq}