To evaluate

$\displaystyle \small \small \small \small \small \int_2^\pi e^{\pi x-1}dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small F(\pi)-F(2)$. With $\displaystyle \small \small \small \small F(x)=\frac{e^{\pi x-1}}{\pi}$, we get

$\displaystyle \small \small \small \int_2^\pi e^{\pi x-1}dx=\frac{1}{\pi}(e^{\pi^2-1}-e^{2\pi-1})$

To evaluate

$\displaystyle \small \small \small \int_1^{e^2} \frac{3}{x}dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small F(e^2)-F(1)$. With $\displaystyle \small \small F(x)=3\ln x$, we get

$\displaystyle \small \small \small \small \int_0^{e^2} \frac{3}{x}dx=3(\ln e^2-\ln 1)=3(2-0)=6$

To evaluate

$\displaystyle \small \small \small \small \int_5^6 x^2dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small F(6)-F(5)$. With $\displaystyle \small \small \small F(x)=\frac{x^3}{3}$, we get

$\displaystyle \small \small \int_5^{6} x^2dx=\frac{1}{3}(6^3-5^3)=\frac{1}{3}(216-125)=\frac{91}{3}$

To evaluate

$\displaystyle \small \small \small \small \small \small \int_2^5 \frac{1}{x}dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small F(5)-F(2)$. With $\displaystyle \small \small \small \small \small F(x)=\ln x$, we get

$\displaystyle \small \small \small \small \small \int_2^5 \frac{1}{x}dx=\ln 5-\ln 2=\ln \frac{5}{2}$

To evaluate

$\displaystyle \small \small \small \int_{-1}^1 5x^5 dx$

we can see that $\displaystyle \small f(x)=5x^5$ is odd (what this means is that $\displaystyle \small f(-x)=-f(x)$), which means that the integral evaluates to $\displaystyle \small 0$, so we get

$\displaystyle \small \small \small \small \int_{-1}^1 5x^5 dx=0$

To evaluate

$\displaystyle \small \small \int_1^3 4x dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small \small F(3)-F(1)$. With $\displaystyle \small \small \small \small \small \small F(x)=2x^2$, we get

$\displaystyle \small \small \int_1^3 4x dx=2(3^2-1^2)=2(9-1)=16$

To evaluate

$\displaystyle \small \small \small \int_0^1 \pi x dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small \small \small F(1)-F(0)$. With $\displaystyle \small \small \small \small \small \small \small F(x)=\frac{\pi}{2}x^2$, we get

$\displaystyle \small \small \small \int_0^1 \pi x dx=\frac{\pi}{2}(1^2-0^2)=\frac{\pi}{2}$

To evaluate

$\displaystyle \small \small \small \small \int_0^1 \frac{3}{x+1} dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small \small \small F(1)-F(0)$. With $\displaystyle \small \small \small \small \small \small \small \small F(x)=3\ln(x+1)$, we get

$\displaystyle \small \small \small \small \small \small \int_0^1 \frac{3}{x+1}dx=3(\ln(1+1)-\ln(0+1))=3\ln 2$

To evaluate

$\displaystyle \small \small \int_0^1 ex^2 dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small \small \small F(1)-F(0)$. With $\displaystyle \small \small F(x)=\frac{e}{3}x^3$, we get

$\displaystyle \small \small \int_0^1 ex^2dx=\frac{e}{3}(1^3-0^3)=\frac{e}{3}$

To evaluate

$\displaystyle \small \small \small \int_0^1 e^2x^3 dx$

we take its antiderivative $\displaystyle \small F(x)$ and calculate $\displaystyle \small \small \small \small \small F(1)-F(0)$. With $\displaystyle \small \small \small F(x)=\frac{e^2}{4}x^4$, we get

$\displaystyle \small \small \small \int_0^1 e^2x^3dx=\frac{e^2}{4}(1^4-0^4)=\frac{e^2}{4}$