Answer to Question #131760 in Calculus for Navya Sharma

Question

evaluate limit 6 to 0 ∫ f(x)dx if f(x){x^2,x<2,3x-2,x>=2}

Answer 1

$\int \limits_0^6 f(x)\,dx=\int \limits_0^2 x^2\,dx+\int \limits_2^6 (3x-2)\,dx=\\=\frac{x^3}3 \bigg|_0^2 +3\frac{x^2}2 \bigg|_2^6 - 2x \bigg|_2^6 =\\= \frac{8}3+\frac{3}2 (36-4) - 2 (6-2) = 42\frac{2}3\\ \int\limits_6^0 f(x)\,dx=-\int\limits_0^6f(x)\,dx=>\int\limits_6^0 f(x)\,dx=-42\frac{2}3$