First slide

Integration

Question

$\int_{0}^{1} (t^{2} + 9 t + c) d t = \frac{9}{2}$ Find the value of 'c'

Easy

A

$- \frac{1}{3}$
B

Zero
C

3
D

2

Solution

$\begin{array}{l} = \int_{0}^{1} t^{2} + 9 t + c) d t = \frac{9}{2} \\ {= \frac{t^{3}}{3} + \frac{9}{2} t^{2} + c t]}_{0}^{1} = \frac{9}{2} \\ = \frac{1}{3} + \frac{9}{2} + c = \frac{9}{2} \\ \Rightarrow c = - \frac{1}{3} \end{array}$

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