Question

question: determine the value of $int_{k}^{2}f(x)dx$ given that $int_{0}^{2}f(x)dx=-7$ and $int_{0}^{k}f(x)dx = 1$. sorry, thats incorrect. try again? feedback view answer submit

Explanation:

Step1: Use integral property

We know that $\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx$. So, $\int_{k}^{2}f(x)dx=-\int_{2}^{k}f(x)dx$. Also, $\int_{2}^{k}f(x)dx=\int_{2}^{9}f(x)dx+\int_{9}^{k}f(x)dx$.

Step2: Substitute given values

Given $\int_{2}^{9}f(x)dx = - 7$ and $\int_{9}^{k}f(x)dx=1$. Then $\int_{2}^{k}f(x)dx=-7 + 1=-6$.

Step3: Find the required integral

Since $\int_{k}^{2}f(x)dx=-\int_{2}^{k}f(x)dx$, substituting the value of $\int_{2}^{k}f(x)dx$ we get $\int_{k}^{2}f(x)dx=-(-6)=6$.

Answer:

$6$