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Question

Mathematics Question on integral

If ∫−32f(x)dx=73\int^{2}_{-3} f\left(x\right)dx = \frac{7}{3} and ∫−39f(x)dx=−56,\int^{9}_{-3} f\left(x\right)dx = - \frac{5}{6} , then what is the value of ∫29f(x)dx\int^{9}_{2} f\left(x\right)dx ?

A

−196 - \frac{19}{6}

B

196 \frac{19}{6}

C

32\frac{3}{2}

D

−32 - \frac{3}{2}

Answer

−196 - \frac{19}{6}

Explanation

Solution

Value of the integral ∫29f(x)dx\int^{9}_2 f(x) dx =∫−39f(x)dx−∫−32f(x)dx= \int^{9}_{-3} f\left(x\right)dx - \int^{2}_{-3} f\left(x\right)dx Given, ∫−39f(x)dx=−56\int^{9}_{-3} f\left(x\right)dx = \frac{-5}{6} and ∫−32f(x)dx=73\int^{2}_{- 3}f\left(x\right)dx = \frac{7}{3} Putting these values in equation (i) ∫29f(x)dx=−56−73=−196\int^{9}_{2} f\left(x\right)dx = \frac{-5}{6} - \frac{7}{3} = - \frac{19}{6}