WAEC 2018 · Paper 2 · Q2

If $\displaystyle\int_1^t (2x + 5)\,dx = 18$ , find the value of $t$ .

(a)
Value of $t$

Worked solution (try it first)

Integrate: $\displaystyle\int (2x + 5)\,dx = x^2 + 5x$ .
Apply the limits: $(t^2 + 5t) - (1^2 + 5 \times 1) = t^2 + 5t - 6$ .
Set it equal to 18: $t^2 + 5t - 6 = 18$ , so $t^2 + 5t - 24 = 0$ .
Factorise: $(t + 8)(t - 3) = 0$ , so $t = 3$ or $t = -8$ .
The upper limit $t$ is meant to be above the lower limit 1, so $t = 3$ .

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