By definition, the integral of a fucntion \(\displaystyle f (x)\) is the summation of an infinite number of small areas, thus giving the total area of the function with respect to the axis of integration.

We have a seperable integral, meaning each term can be integrated independently, written as

\(\displaystyle \int f (x)dx= \int \left(x^4 - \cos (2x)\right )dx = \int x^4 dx - \int \cos(2x)dx\).

The first term is a power-rule integral, while the second is a trigonometric intergral.  One should recall

\(\displaystyle \int x^n dx = \frac{1}{n+1}x^{n+1}+C\)

and \(\displaystyle \int \cos (ax)dx = \frac{1}{a} \sin(ax)+C\)

Putting these facts together leads us to the final answer of

\(\displaystyle \int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)+C\).

To find the position function from the acceleration function, integrate \(\displaystyle a(t)= 2t+11\) twice.

\(\displaystyle v(t)=\int a(t)dt=t^2+11t\)

When integrating, remember to increase the exponent of the variable by one and then divide the term by the new exponent. Do this for each term.

\(\displaystyle s(t)=\int v(t)dt=\int\int a(t)dt = \frac{1}{3}t^3+\frac{11}{2}t^2\)

Solve for \(\displaystyle s(t=1)\).

\(\displaystyle \frac{1}{3}(1)^3+\frac{11}{2}(1)^2=\frac{1}{3}+\frac{11}{2} = \frac{2}{6}+\frac{33}{6}= \frac{35}{6}\)

The correct answer is \(\displaystyle \frac{2}{3}(2^{3/2}-1)\).

We can evaluate this integral using \(\displaystyle u\)-substitution.

Let \(\displaystyle u =1+t^2\), then \(\displaystyle du = 2tdt\), hence we have

\(\displaystyle \int_0^1 2t\sqrt{1+t^2}dt\) (Don't forget to change the bounds of integration)

\(\displaystyle =\int_1^2 \sqrt{u} du\)

\(\displaystyle = \int_1^2 u^{1/2}du\)

\(\displaystyle =[\frac{2u^{3/2}}{3}]^2_1\)

\(\displaystyle =\frac{2\times{2}^{3/2}}{3} -\frac{2}{3}\)

\(\displaystyle =\frac{2}{3}(2^{3/2}-1)\).

Since \(\displaystyle a\) is a constant, so is \(\displaystyle e^a\), therefore we can factor it out of the integral.

\(\displaystyle \int_0^te^ax dx\)

\(\displaystyle =e^a\int_0^txdx\)

\(\displaystyle =e^a[\frac{x^2}{2}]^t_0\)

\(\displaystyle =e^a(\frac{t^2}{2}-0)\)

\(\displaystyle =\frac{t^2e^a}{2}\)

This integral can be found using u-substitution.  Consider

\(\displaystyle u = \sin(x), du = \cos(x)\).  This means we can rewrite our integral as

\(\displaystyle \int e^{u}du=e^u+C\), by definition of the integral of an exponential.

This integral can be done using integration by parts.  Consider

\(\displaystyle \int 1 \cdot \ln x dx\).

Choose \(\displaystyle u = \ln x, du = \frac{dx}{x}\) and \(\displaystyle dv = dx, v = x\).

Using the definition of integration by parts,

\(\displaystyle \int u dv = uv - \int v du\),

\(\displaystyle uv - \int v du= x \ln x - \int dx= x \ln x - x\).

This can be a challenging integral using standard methods.  However, it is easy if we use integration by-parts, given as

\(\displaystyle \int udv = uv - \int v du\)

Choose

\(\displaystyle u = x, dv = e^{-x} \Rightarrow du = dx, v = -e^{-x}\).

From the definition,

\(\displaystyle 2\int _0^{\infty}xe^{-x}dx= 2\left(-x*e^{-x} \left | _0 ^{\infty}+\int _0^{\infty}e^{-x}dx\right)=2\left(-e^{-x} \left| _0^{\infty}\right) = 2\)

This integral is most easily found by implementing u-substitution.  Choose 

\(\displaystyle u = 2 \sin \theta + 5 \Rightarrow \frac{du}{2} = \cos \theta d \theta\), which means we can rewrite the integral in a more familiar form

\(\displaystyle \int \cos \theta \left( 2 \sin \theta + 5\right )^3 d \theta = \frac{1}{2}\int u^3 du= \frac{1}{2}*\frac{1}{4}u^4+C= \frac{1}{8}\left(2 \sin \theta + 5 \right )^4 + C\)

This integral is most easily done by using u-substitution.  Initially rewrite our integral as

\(\displaystyle \int \frac{6x+15}{\left(x^2+5x+10 \right )^3} dx=\int \frac{3(2x+5)}{\left(x^2+5x+10 \right )^3} dx\), then choose

\(\displaystyle u = x^2 + 5x + 10, du = (2x+5)dx\).  Therefore

\(\displaystyle \int \frac{6x+15}{\left(x^2+5x+10 \right )^3} dx= 3\int u^{-3}du= -\frac{3}{2}u^{-2}+C=-\frac{3}{2\left(x^2+5x+10 \right )^2}+C\)