Đáp án:
Câu 14:
Đặt $x=-t.$ Khi đó:
$\displaystyle\int\limits^a_{-a}f(x)dx=\displaystyle\int\limits^{-a}_af(-t)(-dt)=\displaystyle\int\limits^a_{-a}f(-t)dt=\displaystyle\int\limits^a_{-a}f(-x)dx$
Do đó: $2\displaystyle\int\limits^a_{-a}f(x)dx=\displaystyle\int\limits^a_{-a}[f(x)+f(-x)]dx=\displaystyle\int\limits^a_{-a}adx\\\to 2\displaystyle\int\limits^a_{-a}f(x)dx=2a^2\\\to \displaystyle\int\limits^a_{-a}f(x)dx=a^2$
$\to C$
Câu 18:
Gọi $F(x)=\displaystyle\int\sqrt{1+x^2}dx\to F'(x)=\sqrt{1+x^2}$ và $G(t)=\displaystyle\int\limits^t_1\sqrt{1+x^2}dx=F(t)-F(1)$
$\to G'(t)=F'(t)=\sqrt{1+t^2}$
$\to D$
Câu 19: Sử dụng tích phân từng phần, ta có:
$\displaystyle\int\limits^2_1F(x)g(x)dx=$
$[F(x)G(x)]\Big|^2_1-\displaystyle\int\limits^2_1f(x)G(x)dx\\=F(2)G(2)-F(1)G(1)-\displaystyle\int\limits^2_1f(x)G(x)dx\\=4\cdot 2-\dfrac 32\cdot 1-\dfrac{67}{12}\\=\dfrac{11}{12}\\\to A$
