# Compute the following: integral < 4 t^3, t e^{t^2} > dt.

## Question:

Compute the following:

{eq}\displaystyle \int \langle 4 t^3,\ t e^{t^2} \rangle\ dt {/eq}.

## Integral of a function

Integral is used to determine the area under the curve or graph. It is also known as antiderivative of a function. Two integral formulas are used to solve the given problem and are as follows:

{eq}1. \displaystyle \int {{z^n}} dz = \frac{{{z^{n + 1}}}}{{n + 1}} + C\\ 2. \displaystyle \int {{e^z}} dz = {e^z} + C \\ {/eq}

where {eq}C {/eq} is integration constant.

## Answer and Explanation:

Given problem is

{eq}\displaystyle \int {\langle 4{t^3},\;t{e^{{t^2}}}\rangle } \;dt......(i)\\ \displaystyle = \int {\langle 4{t^3}\rangle } \;dt,\int {\langle \;t{e^{{t^2}}}\rangle } \;dt.....(ii) {/eq}

Taking left part of the equation (ii)

{eq}\displaystyle \int {\langle 4{t^3}\rangle } \;dt\\ \displaystyle = 4\int {\langle {t^3}\rangle } \;dt {/eq}

Using power rule {eq}\displaystyle \int {{z^n}} dz = \frac{{{z^{n + 1}}}}{{n + 1}} + C {/eq}

{eq}\displaystyle = 4\frac{{{t^{3 + 1}}}}{{3 + 1}}\\ \displaystyle = \langle {t^4} + C1\rangle ....(iii) {/eq}

Taking the right side of the equation (ii)

{eq}\displaystyle \int {\langle \;t{e^{{t^2}}}\rangle } \;dt {/eq}

Substituting {eq}\displaystyle u = {t^2}.....(iv) \to \frac{{du}}{{dt}} = 2t \to dt = \frac{1}{{2t}}du {/eq}, we get

{eq}\displaystyle = \frac{1}{2}\int {{e^u}} du {/eq}

Using integral formula {eq}\displaystyle \int {{e^z}} dz = {e^z} + C2 {/eq}

{eq}\displaystyle = \frac{1}{2}{e^u} + C2 {/eq}

Undo substitution by putting the value of u from equation (iv)

{eq}\displaystyle = \frac{1}{2}{e^{{t^2}}} + C2....(v) {/eq}

Putting the values from equation (iii) and (v) into equation (i)

{eq}\displaystyle = \langle {t^4} + C1,\frac{1}{2}{e^{{t^2}}} + C2\rangle {/eq}

where C1 and C2 are the integration constants.

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13