$\displaystyle \int_{a}^{b} f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

In this case, $\displaystyle f(x) = \sin x$.

The antiderivative is  $\displaystyle F(x) = - \cos x$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{0}^{1} \sin x \ dx = \left ( -\cos x )_{0}^{1}$

$\displaystyle = - \cos(1) - \left ( - \cos (0) \right )$

$\displaystyle = 1.54$

By a straightforward application of antiderivative rules for polynomial functions,

$\displaystyle \int_{-3}^{3}(1+x^2)dx$ $\displaystyle = \left.\ x+\frac{1}{3}x^3\right|_-_3^3$

                            $\displaystyle =(3+9)-(-3-9)$

                            $\displaystyle =12+3+9$

                            $\displaystyle =24$

This integral is tricky, because there doesn't seem to be an apparent substitution method to use. The derivative of $\displaystyle \sec(x)$, $\displaystyle \sec^2(x)$, and $\displaystyle \sec^3(x)$ seem to only complicate the integral further. However, we do note that the integral of $\displaystyle \sec^2(x)$ is given by $\displaystyle \tan(x)$. Thus, we can use integration by parts here. Splitting $\displaystyle \int\sec^3(x)dx$ into $\displaystyle \int\big(\sec(x)\big)\big(\sec^2(x)dx\big)$ and letting $\displaystyle u=\sec(x)$ and $\displaystyle dv=sec^2(x)dx$, this gives us $\displaystyle du=\sec(x)\tan(x)dx$ and $\displaystyle v=\tan(x)$. Plugging this into the integration by parts formula, we get:

$\displaystyle \int\sec^3(x)dx=\int\big(\sec(x)\big)\big(\sec^2(x)dx\big)=\int udv=uv-\int vdu$

$\displaystyle \int\sec^3(x)dx=\big(\sec(x)\big)\big(\tan(x)\big)-\int\big(\tan(x)\big)\big(\sec(x)\tan(x)dx\big)$

$\displaystyle \int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec(x)\tan^2(x)dx$

Unfortunately, this still looks more complicated than the original. However, using the trig identity $\displaystyle \tan^2(x)=\sec^2(x)-1$ on the second integral will give us something we can work with. Doing so we obtain:

$\displaystyle \int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec(x)\big(\sec^2(x)-1\big)dx$

$\displaystyle \int\sec^3(x)dx=\sec(x)\tan(x)-\int\big(\sec^3(x)-\sec(x)\big)dx$

$\displaystyle \int\sec^3(x)dx=\sec(x)\tan(x)-\int\sec^3(x)dx+\int\sec(x)dx$

Now here, we know how to evaluate the integral of $\displaystyle \sec(x)$, and we also have a copy of our integral $\displaystyle \int\sec^3(x)dx$ subtracted on the right side. Therefore, we can add this integral to each side of the equation, eliminating it on the right side and leaving something we can evaluate. Doing so, we obtain:

$\displaystyle 2\int\sec^3(x)dx=\sec(x)\tan(x)+\int\sec(x)dx$

$\displaystyle 2\int\sec^3(x)dx=\sec(x)\tan(x)+\ln\big(\sec(x)+\tan(x)\big)+D$

Finally, dividing the whole equation by $\displaystyle 2$ and replacing the $\displaystyle \frac{D}{2}$ constant with a new constant $\displaystyle C$ gives us our final result:

$\displaystyle \int\sec^3(x)dx=\frac{1}{2}\biggr(\sec(x)\tan(x)+\ln\big(\sec(x)+\tan(x)\big)\biggr)+C$

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{0}^{1} x^4 \ dx = \left ( \frac{1}{5}x^5 \right )_{0}^{1}$

$\displaystyle = \frac{1}{5}(1)^5 - \left ( \frac{1}{5}(0)^5 \right )$

$\displaystyle = \frac{1}{5}$

The antiderivative of $\displaystyle e^x$ is $\displaystyle e^x$.  Now, all we need to do is plug in our bounds, remembering it is always top - bottom bound.

$\displaystyle \int_{-2}^{4}e^xdx=e^4-\frac{1}{e^2}.$  In the last step, I simply rewrote $\displaystyle -e^{-2}$ into the denominator, getting rid of the negative exponent. 

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{0}^{\pi} \cos(2x ) \ dx = \left ( \cos(2x) \right )_{0}^{\pi}$

$\displaystyle = \cos(2(\pi)) - \left ( \cos(2(0)) \right )$

$\displaystyle = 0$

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{2}^{6} \sqrt{x-2} \ dx = \left ( \frac{2(x-2)^{\frac{3}{2}}}{3} \right )_{2}^{6}$

$\displaystyle = \frac{2(6-2)^{\frac{3}{2}}}{3} - \left ( \frac{2(2-2)^{\frac{3}{2}}}{3} \right )$

$\displaystyle = \frac{16}{3}$

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2}e^{3x-1} \ dx = \left ( \frac{e^{3x-1}}{3} \right )_{1}^{2}$

$\displaystyle = \frac{e^{3(2)-1}}{3} - \left ( \frac{e^{3(1)-1}}{3} \right )$

$\displaystyle = 141.02$

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2} 2x + \frac{1}{x} \ dx = \left ( x^2 + \ln(x) \right )_{1}^{2}$

$\displaystyle = 2^2 + \ln(2) - \left ( 1^2 + \ln(1) \right )$

$\displaystyle = 3.69$

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{0}^{1} \cos (2x-3) \ dx = \left ( \frac{\sin(2x-3)}{2} \right )_{0}^{1}$

$\displaystyle = \frac{\sin(2(1)-3)}{2} - \left ( \frac{\sin(2(0)-3)}{2} \right )$

$\displaystyle = -0.35$