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 \(1 - x = 0 ⇔ x = 1\).
\(\int_0^2 {\left| {1 - x} \right|} dx = \int_0^1 {\left| {1 - x} \right|} dx + \int_1^2 {\left| {1 - x} \right|} dx\)
\(= - \int_0^1 {(1 - x)} d(1 - x) + \int_1^2 {(x - 1)} d(x - 1)\)
\( = - {{{{(1 - x)}^2}} \over 2}|_0^1 + {{{{(x - 1)}^2}} \over 2}|_1^2 = {1 \over 2} + {1 \over 2} = 1\)
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b) \(\int_0^{{\pi \over 2}} s i{n^2}xdx\)
\( = {1 \over 2}\int_0^{{\pi \over 2}} {(1 - cos2x)} dx\)
\( = {1 \over 2}\left( {x - {1 \over 2}sin2x} \right)|_0^{{\pi \over 2}} = {\pi \over 4}\)
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 : \(\eqalign{
& \int_0^\pi s in2xco{s^2}xdx = \int_0^\pi {({1 \over 2}sin2x + {1 \over 4}sin4x)} dx \cr
& = ( - {1 \over 4}cos2x - {1 \over {16}}cos4x)|_0^\pi \cr
& = - {1 \over 4} - {1 \over {16}} + {1 \over 4} + {1 \over {16}} = 0 \cr} \).