Integration Bee

2500

Vi har att
\begin{align*}
\int_1^{11}(x^3-3x^2+3x-1)\, \diff{x}
&
=\int_1^{11}\bigl(x^2(x-3)+3x-1\bigr)\, \diff{x}
\\&
\qquad \text{\(t=x-1 \Leftrightarrow x=t+1\), \(\tfrac{ \diff{t}}{ \diff{x}}=1\)}
\\&
=\int_0^{10}\bigl((t+1)^2\bigl((t+1)-3\bigr)+3(t+1)-1\bigr)\, \diff{t}
\\&
=\int_0^{10}\bigl((t^2+2t+1)(t-2)+3t+2\bigr)\, \diff{t}
\\&
=\int_0^{10}(t^3-2t^2+2t^2-4t+t-2+3t+2)\, \diff{t}
\\&
=\int_0^{10}t^3\, \diff{t}
=\bigl[\tfrac{1}{4}t^4\bigr]_0^{10}
=\tfrac{1}{4}\bigl[t^4\bigr]_0^{10}
=\tfrac{1}{4}(10^4-0^4)
=\tfrac{1}{4}\cdot10^4
=2500.
\end{align*}

\(\sqrt{3} \pi\)

Vi har att
\begin{align*}
\int_0^2\sqrt{12-3x^2}\, \diff{x}
&
=\int_0^2\sqrt{3(4-x^2)}\, \diff{x}
=\int_0^2\sqrt{3}\cdot\sqrt{2^2-x^2}\, \diff{x}
=\sqrt{3}\int_0^2\sqrt{2^2-x^2}\, \diff{x}.
\end{align*}
Grafen \(y=\sqrt{2^2-x^2}\), \(x\in[-2,2]\), beskriver en halvcirkel med radien 2 och medelpunkt i origo varför \(\int_0^2\sqrt{2^2-x^2}\, \diff{x}\) är arean av halva halvcirkeln med radien 2, d.v.s.
\[
\int_0^2\sqrt{2^2-x^2}\, \diff{x}=\tfrac{1}{2}\cdot\tfrac{1}{2}\cdot \pi\cdot2^2= \pi
\]
varför
\[
\int_0^2\sqrt{12-3x^2}\, \diff{x}=\sqrt{3} \pi.
\]

18

Vi har att
\begin{align*}
\int_0^6\bigl(x+(x-3)^7+\sin(x-3)\bigr)\, \diff{x}
&
=\int_0^6x\, \diff{x}+\int_0^6(x-3)^7\, \diff{x}+\int_0^6\sin(x-3)\, \diff{x}
\\&
\quad\text{\(t=x-3\), \( \diff{t}= \diff{x}\)}
\\&
=\bigl[\tfrac{1}{2}x^2\bigr]_0^6+\int_{-3}^3t^7\, \diff{t}+\int_{-3}^3\sin(t)\, \diff{t}
\\&
\quad\text{\(t^7\) och \(\sin(t)\) är udda funktioner omkring origo}
\\&
=\tfrac{1}{2}\bigl[x^2\bigr]_0^6+0+0
=\tfrac{1}{2}(6^2-0^2)
=18.
\end{align*}