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Bài 2. Tính các tích phân sau:
a) \(\int_0^2 {\left| {1 – x} \right|} dx\) b) \(\int_0^{{\pi \over 2}} s i{n^2}xdx\)
c) \(\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx\) d) \(\int_0^\pi s in2xco{s^2}xdx\)
Hướng dẫn giải:
a) Ta có \(1 – x = 0 ⇔ x = 1\).
\(\int_0^2 {\left| {1 – x} \right|} dx = \int_0^1 {\left| {1 – x} \right|} dx + \int_1^2 {\left| {1 – x} \right|} dx\)
\(= – \int_0^1 {(1 – x)} d(1 – x) + \int_1^2 {(x – 1)} d(x – 1)\)
\( = – {{{{(1 – x)}^2}} \over 2}|_0^1 + {{{{(x – 1)}^2}} \over 2}|_1^2 = {1 \over 2} + {1 \over 2} = 1\)
b) \(\int_0^{{\pi \over 2}} s i{n^2}xdx\)
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\( = {1 \over 2}\int_0^{{\pi \over 2}} {(1 – cos2x)} dx\)
\( = {1 \over 2}\left( {x – {1 \over 2}sin2x} \right)|_0^{{\pi \over 2}} = {\pi \over 4}\)
c) \(\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx = \int_0^{ln2} {({e^{x + 1}} + {e^{ – x}})} dx\)
\( = ({e^{x + 1}} – {e^{ – x}})|_0^{ln2} = e + {1 \over 2}\)
d) Ta có : \(sin2xcos^2x\) = \({1 \over 2}sin2x(1 + cos2x) = {1 \over 2}sin2x + {1 \over 4}sin4x\)
Do đó : \(\eqalign{
& \int_0^\pi s in2xco{s^2}xdx = \int_0^\pi {({1 \over 2}sin2x + {1 \over 4}sin4x)} dx \cr
& = ( – {1 \over 4}cos2x – {1 \over {16}}cos4x)|_0^\pi \cr
& = – {1 \over 4} – {1 \over {16}} + {1 \over 4} + {1 \over {16}} = 0 \cr} \).