Giải thích các bước giải:
Ta có :$I=\int^1_{-1}f(x)dx$
$\to I=\int^{-1}_1f(-x)d(-x)=-\int^{-1}_1f(-x)dx=\int^1_{-1}f(-x)dx$$\to 2I=\int^1_{-1}f(x)+f(-x)dx$
$\to 2I=\int^1_{-1}x^2dx=\dfrac{x^3}{3}\Big|^1_{-1}=\dfrac 23$
$\to I=\dfrac 13\to D$
Đáp án:
I=∫1−1f(x)dxI=∫−11f(x)dx
→I=∫−11f(−x)d(−x)=−∫−11f(−x)dx=∫1−1f(−x)dx→I=∫1−1f(−x)d(−x)=−∫1−1f(−x)dx=∫−11f(−x)dx→2I=∫1−1f(x)+f(−x)dx→2I=∫−11f(x)+f(−x)dx
→2I=∫1−1x2dx=x33∣∣1−1=23→2I=∫−11x2dx=x33|−11=23
→I=13→D
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