Cho ∫0π2f(x)dx=5{\int\limits_{0}^{\dfrac{\pi}{2}}{f(x)\text{d}x}} = 50∫2πf(x)dx=5. Tính I=∫0π2[f(x)+2sinx]dxI = {\int\limits_{0}^{\dfrac{\pi}{2}}{\left\lbrack {f(x) + 2\sin x} \right\rbrack\text{d}x}}I=0∫2π[f(x)+2sinx]dx?
I=7I = 7I=7.
I=5+π2I = 5 + \dfrac{\pi}{2}I=5+2π.
I=3I = 3I=3.
I=5+πI = 5 + \piI=5+π.
Đáp án: <p><span data-type="math-inline" data-latex="I = 7">$I = 7$</span>.</p>
Phương pháp giải:
Áp dụng tính chất của tích phân ∫ab[f(x)+g(x)]dx=∫abf(x)dx+∫abg(x)dx\int\limits_a^b {\left[ {f(x) + g(x)} \right]dx} = \int\limits_a^b {f(x)dx} + \int\limits_a^b {g(x)dx}a∫b[f(x)+g(x)]dx=a∫bf(x)dx+a∫bg(x)dx.
Lời giải chi tiết:
I=∫0π2[f(x) +2sinx] dx=∫0π2f(x) dx +2∫0π2sinxdxI = \int\limits_0^{\frac{\pi }{2}} {\left[ {f\left( x \right)\, + 2\sin x} \right]\,{\rm{d}}x = \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)\,{\rm{d}}x\,{\rm{ + 2}}\int\limits_0^{\frac{\pi }{2}} {\sin x} dx} }I=0∫2π[f(x)+2sinx]dx=0∫2πf(x)dx+20∫2πsinxdx=∫0π2f(x) dx −2cosx∣π20=5−2(0−1)=7= \int\limits_0^{\frac{\pi }{2}} {f\left( x \right)} \,{\rm{d}}x\,\, - 2\cos x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. = 5 - 2\left( {0 - 1} \right) = 7=0∫2πf(x)dx−2cosx2π0=5−2(0−1)=7.
