Definite Integration Question - None

Question

For $c < 1$, find the value of $c$ such that $\int_c^1 (8x^2 - x^5) \, dx = \frac{16}{3}$.

Accepted Answer

Step 1: Compute the indefinite integral: $\int (8x^2 - x^5) \, dx = \frac{8x^3}{3} - \frac{x^6}{6}$

Step 2: Apply limits from $c$ to $1$:

$\left[\frac{8x^3}{3} - \frac{x^6}{6}\right]_c^1 = \frac{8}{3} - \frac{1}{6} - \left(\frac{8c^3}{3} - \frac{c^6}{6}\right) = \frac{16}{3}$

Step 3: Simplify: $\frac{8}{3} - \frac{1}{6} - \frac{8c^3}{3} + \frac{c^6}{6} = \frac{16}{3}$

Step 4: Solve for $c$: $c = -1$ satisfies the equation for $c < 1$.

∴ Answer is $c = -1$.

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