## H1011 Quizzes

Quiz 11: One variable integration
Question 1 Questions
Evaluate $∫ 01(6x2 − 4x + 3)dx$.
 a) 4 b) -3 c) 3 d) 5

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$∫ 01(6x2 − 4x + 3)dx = 2x3 − 2x2 + 3x 01 = 2 − 2 + 3 = 3.$
Choice (d) is incorrect
Evaluate $∫ 0π 2 (sinx + cosx)dx$.
 a) 2 b) 0 c) $π 2$ d) -2

Choice (a) is correct!
$∫ 0π 2 (sinx + cosx)dx = −cosx + sinx0π 2 = −cos π 2 + sin π 2 − (−cos0 + sin0) = 1 + 1 = 2.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate $∫ 0π 4 sin2xdx$.
 a) 1 b) $1 2$ c) $−1 2$ d) -1

Choice (a) is incorrect
Choice (b) is correct!
$∫ 0π 4 sin2xdx = −1 2cos2x0π 4 = −1 2cos π 2 + 1 2cos0 = 1 2.$
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate $∫ 03x 1 + x2 3 2 dx$.
 a) $32 5$ b) $31 5$ c) $3 2$ d) $9 2$

Choice (a) is incorrect
Choice (b) is correct!
$∫ 03x 1 + x2 3 2 dx = 1 5(1 + x2)5 2 03 = 1 5 45 2 −1 5 = 32 5 −1 5 = 31 5 .$
Choice (c) is incorrect
Choice (d) is incorrect
Evaluate $∫ 023x2ex3 dx$.
 a) $e8 − 1$ b) $8e8$ c) $−e8$ d) $12e8 − 12$

Choice (a) is correct!
$∫ 023x2ex3 dx = ex3 02 = e8 − e0 = e8 − 1.$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Let $A$ be the area under the curve $f(x) = 16 − x2$ on the interval [0,4]. Dividing the interval into 4 subintervals and defining $xi$ to be the midpoint of the $i$th interval (the area of the rectangle is thus $f(xi)$ times the length of the interval), the best estimate for $A$ is
 a) $A = ∑ i=14(16 − x i2) × 1 2$, where $xi = i − 1 2$ b) $A = ∑ i=14(16 − x i2) × 1 2$, where $xi = i − 1 2$ c) $A = ∑ i=14(16 − x i2)$, where $xi = 2i − 1 2$ d) $A = ∑ i=03(16 − x i2)$, where $xi = 2i − 1 2$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The intervals are of length 1, and the area is calculated at $1 2$, $3 2$, $5 2$ and $7 2$.
Choice (d) is incorrect
Which of the following sums is the best estimate for $A$, the area under the curve $f(x) = x3 + 2$ on the interval [-1,2], divided into 6 subintervals and choosing $xi$ as the right-endpoint of the $i$th interval ?
 a) $A = ∑ i=06(x i3 + 2)$, where $xi = i − 3 2$ b) $A = ∑ i=06(x i3 + 2) × 1 2$, where $xi = 2i − 1 2$ c) $A = ∑ i=16(x i3 + 2)$, where $xi = 2i − 3 2$ d) $A = ∑ i=16(x i3 + 2) × 1 2$, where $xi = i − 2 2$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
The intervals are of length $1 2$, and the area is calculated at $−1 2$, $0$, $1 2$, $1$, $3 2$, $2$
Find the indefinite integral $∫ 2t2(1 + t3)4dt.$
 a) $1 5(1 + t3)5 + C$ b) $2 5(1 + t3)5 + C$ c) $2 3t3(1 + t3)5 + C$ d) $2 15(1 + t3)5 + C$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
Let $u = 1 + t3, du = 3t2dt$. $∫ 2t2(1 + t3)4dt = 2 3∫ u4du = 2 15u5 + C = 2 15(1 + t3)5 + C.$
Find the indefinite integral $∫ sin3xcosxdx.$
 a) $sin4x + C$ b) $1 4sin4x + C$ c) $3sin2x + C$ d) $−1 4sin4x + C$

Choice (a) is incorrect
Choice (b) is correct!
Let $u = sinx$, $du = cosxdx.$ $∫ sin3xcosxdx = ∫ u3du = 1 4u4 + C = 1 4sin4x + C.$
Choice (c) is incorrect
Choice (d) is incorrect
Find the indefinite integral
$∫ 2t − e−t + 4 (t2 + 4t + e−t + 1)2dt.$
 a) $1 t2 + 4t + e−t + 1 + C$ b) $−1 3(t2 + 4t + e−t + 1)3 + C$ c) $−1 t2 + 4t + e−t + 1 + C$ d) $1 3(t2 + 4t + e−t + 1)3 + C$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Let $u = t2 + 4t + e−t + 1$, $du = (2t + 4 − e−t)dt.$ $∫ 2t − e−t + 4 (t2 + 4t + e−t + 1)2dt = ∫ 1 u2du = −u−1 + C = − 1 t2 + 4t + e−t + 1 + C.$
Choice (d) is incorrect