P.o.t.W. #26
Problem of the Week #26
■ No GDC ■
for HL students
(a) (i) Show that \(\displaystyle\int_0^a {{x^2}{\textrm{d}}x} = \frac{1}{3}{a^3}\)
(ii) Find \(\displaystyle\frac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {\frac{1}{3}{a^3}} \right)\)
(b) (i) Show that \(\displaystyle\int_0^a {\cos \left( x \right)\,{\textrm{d}}x} = \sin \left( a \right)\)
(ii) Find \(\dfrac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {\sin \left( a \right)} \right)\)
(c) (i) Show that \(\displaystyle\int_0^a {\sqrt x \,{\textrm{d}}x} = \frac{2}{3}\sqrt {{a^3}} \)
(ii) Find \(\dfrac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {\dfrac{2}{3}\sqrt {{a^3}} } \right)\)
(d) (i) Show that \(\displaystyle\int_0^a {{{\textrm{e}}^x}{\textrm{d}}x} = {{\textrm{e}}^a} - 1\)
(ii) Find \(\dfrac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {{{\textrm{e}}^a} - 1} \right)\)
(e) (i) Show that \(\displaystyle\int_0^a {\sin \left( x \right)\,{\textrm{d}}x} = 1 - \cos \left( a \right)\)
(ii) Find \(\dfrac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {1 - \cos \left( a \right)} \right)\)
(f) (i) Using integration by parts, show that \(\displaystyle\int_0^a {\ln \left( x \right)\,{\textrm{d}}x} = a\ln \left( a \right) - a\)
(ii) Find \(\dfrac{{\textrm{d}}}{{{\textrm{d}}a}}\left( {a\ln \left( a \right) - a} \right)\)
(g) Using your results from (a) - (f), complete the following statement:
\(\displaystyle\frac{{\textrm{d}}}{{{\textrm{d}}a}}\left[ {\int_0^a {f\left( x \right)\,{\textrm{d}}x} } \right] = \)
(h) A solid is generated by rotating about the x-axis the region under the curve \(y = f\left( x \right)\) from \(x = 0\) to \(x = b\). The graph of \(f\) is above the x-axis for \(x \ge 0\). The volume of the solid is \({b^2}\) for all \(b > 0\). Use your statement from (g) to find the function \(f\).
PDF: P.o.t.W. #26