Calculus Volume 1

Chapter 5 | Integration

593

⌠ ⌡ ⎮ y 5 ⎛ ⎝ 1− y 3

π /4

297. ⌠

sin θ cos 4 θ

⌡ 0

dy

283.

3/2

⎞ ⎠

In the following exercises, evaluate the indefinite integral ∫ f ( x ) dx with constant C =0 using u -substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of C that would need to be added to the antiderivative to make it equal to the definite integral F ( x ) = ∫ a x f ( t ) dt , with a the left endpoint of the given interval. 298. [T] ∫ (2 x +1) e x 2 + x −6 dx over [−3, 2] 299. [T] ∫ cos ⎛ ⎝ ln(2 x ) ⎞ ⎠ x dx on [0, 2]

284. ∫ cos θ (1−cos θ ) 99

sin θdθ

10

⎛ ⎝ 1−cos 3 θ ⎞ ⎠

285. ∫ 286. ⌠

cos 2 θ sin θdθ

⎝ cos 2 θ −2cos θ ⎞ ⎠ 3

⌡ (cos θ −1) ⎛

sin θdθ

287. ⌠ ⌡ ⎛

⎝ sin 3 θ −3sin 2 θ ⎞ ⎠ 3

⎝ sin 2 θ −2sin θ ⎞ ⎠ ⎛

cos θdθ

In the following exercises, use a calculator to estimate the area under the curve using left Riemann sums with 50 terms, then use substitution to solve for the exact answer. 288. [T] y =3(1− x ) 2 over [0, 2]

300. [T] ⌠

x 2 +2 x +1 x 3 + x 2 + x +4

⌡ 3

dx over [−1, 2]

301. [T] ⌠ ⌡

⎡ ⎣ − π

⎤ ⎦

π 3

sin x cos 3 x

3

dx over

289. [T] y = x ⎛

⎞ ⎠

⎝ 1− x 2

3 ,

over [−1, 2]

2 −4 x +3

290. [T] y = sin x (1−cos x ) 2 over [0, π ] 291. [T] y = x ⎛ ⎝ x 2 +1 ⎞ ⎠ 2 over [−1, 1]

302. [T] ∫ ( x +2) e − x

dx over ⎡

⎣ −5, 1 ⎤ ⎦

303. [T] ∫ 3 x 2 2 x 3 +1 dx over [0, 1] 304. If h ( a ) = h ( b ) in ∫ a b g ' ⎛ ⎝ h ( x ) ⎞

In the following exercises, use a change of variables to evaluate the definite integral. 292. ∫ 0 1 x 1− x 2 dx 293. ⌠ ⌡ 0 1 x 1+ x 2 dx 294. ⌠ ⌡ 0 2 t 2 5+ t 2 dt 295. ⌠ ⌡ 0 1 t 2 1+ t 3 dt 296. ∫ 0 π /4 sec 2 θ tan θdθ

⎠ h ( x ) dx , what can you

say about the value of the integral? 305. Is the substitution u =1− x 2 in the definite integral ⌠ ⌡ 0 2 x 1− x 2 dx okay? If not, why not? In the following exercises, use a change of variables to show that each definite integral is equal to zero. 306. ∫ 0 π cos 2 (2 θ )sin(2 θ ) dθ 307. ∫ 0 π t cos ⎛ ⎝ t 2 ⎞ ⎠ sin ⎛ ⎝ t 2 ⎞ ⎠ dt 308. ∫ 0 1 (1−2 t ) dt

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