Practice with Deﬁnite Integrals Use antidiﬀerentiation to compute the following deﬁnite integrals. Check your work using the geometric deﬁnition of the deﬁnite integral, graphing and esti mation. � 2 a) x2 dx 0 � e b) 1 � 1 dx x 0 c) sin x dx −π/4 Solution � 2 a) x2 dx 0 The antiderivative of x2 is 13 x3 (plus a constant), so: � �2 1 3 �� x 3 �0 1 3 1 3 ·2 − ·0 3 3 8 . 3 2 2 x dx = 0 = = 6 5 4 3 2 1 0 −2 −1 0 1 2 Geometrically, the area under the curve appears to be between 2 and 4 so our answer seems to be correct. � e b) 1 1 dx x The antiderivative of 1 x is ln |x|. � e 1 dx x 1 e = [ln |x|]1 = ln e − ln 1 = = 1−0 1 3 2 1 0 0 1 2 3 The area under the curve does appear to be between 1 and 1.5. � 0 c) sin x dx −π/4 The antiderivative of sin x is − cos x. � 0 0 sin x dx = − cos x|−π/4 −π/4 = = − cos(0) − (− cos(−π/4)) 1 −1 + √ 2 √ 1− 2 = 2 ≈ −0.2 0.5 0 −1.2 −0.8 −0.4 0 0.4 0.8 −0.5 We’re computing the area of a region that lies below the x-axis, so we expect the answer to be negative. The estimate −0.2 seems very small but is in fact close to the correct value, as we see when we compare the shaded region to a rectangle with side lengths 0.5 and 0.4.