area under a curve

The area under a curve of above equation is $\displaystyle\int_2^5(x^2+4x-3)\ \mathrm dx$

From,

$\begin{aligned}\displaystyle\int(\color{#D61F06}{x^2}+\color{#20A900}{4x}-\color{#3D99F6}3)\ \mathrm dx=&\ \color{#D61F06}{\displaystyle\int x^2\ \mathrm dx}+\color{#20A900}{4\displaystyle\int x\ \mathrm dx}-\color{#3D99F6}{\displaystyle\int3\ \mathrm dx}\\=&\ \color{#D61F06}{\dfrac{x^3}3}+\color{#20A900}{4\left(\dfrac{x^2}2\right)}-\color{#3D99F6}{3x}+c\\=&\ \color{#D61F06}{\dfrac{x^3}3}+\color{#20A900}{2x^2}-\color{#3D99F6}{3x}+c\end{aligned}$

Thus,

$\begin{aligned}\displaystyle\int_\color{magenta}2^{\color{#302B94}5}(\color{#D61F06}{x^2}+\color{#20A900}{4x}-\color{#3D99F6}3)\ \mathrm dx=&\ \left.\left(\color{#D61F06}{\dfrac{x^3}3}+\color{#20A900}{2x^2}-\color{#3D99F6}{3x}\right)\right|_{\color{magenta}2}^{\color{#302B94}5}\\=&\ \left(\dfrac{\color{#302B94}5^{\color{#D61F06}3}}{\color{#D61F06}3}+\color{#20A900}2\color{#302B94}{(5)}^{\color{#20A900}2}-\color{#3D99F6}3\color{#302B94}{(5)}\right)-\left(\dfrac{\color{magenta}2^{\color{#D61F06}3}}{\color{#D61F06}3}+\color{#20A900}2\color{magenta}{(2)}^{\color{#20A900}2}-\color{#3D99F6}3\color{magenta}{(2)}\right)\\=&\ \dfrac{125}3+50-15-\dfrac83-8+6\\=&\ 39+33=\boxed{72}\end{aligned}$