The rate at which a specific chemical compound is produced during a reaction, in grams per hour, is modeled by the function
R(t)=(4t+1)2t+1 R(t) = (4t + 1)\sqrt{2t + 1} R(t)=(4t+1)2t+1where t t\,t is the time in hours, 0≤t≤50 \le t \le 50≤t≤5.
Use the substitution u=2t+1u = 2t + 1u=2t+1 to show that the total mass of the compound produced,
∫05(4t+1)2t+1 dt \int_{0}^{5} (4t + 1)\sqrt{2t + 1} \, dt ∫05(4t+1)2t+1dtcan be written as
12∫111(2u−1)u12 du=12∫k11(2u32−u12) du \frac{1}{2} \int_{1}^{11} (2u - 1)u^{\frac{1}{2}} \, du = \frac{1}{2} \int_{k}^{11} (2u^{\frac{3}{2}} - u^{\frac{1}{2}}) \, du 21∫111(2u−1)u21du=21∫k11(2u23−u21)duwhere k k\,k is a constant to be found.
Hence, or otherwise, show that the total mass of the compound produced in the first 5 hours is
115(67111−1) grams \frac{1}{15}(671\sqrt{11} - 1) \text{ grams} 151(67111−1) gramsA technician approximates the total mass produced between t=0t = 0t=0 and t=5t = 5t=5 using five rectangles of equal width, where the left-hand edge of each rectangle touches the curve y=R(t)y = R(t)y=R(t). The total area of these five rectangles is MMM.
The technician then decides to use ten rectangles of equal width, still using the left-hand edge method, to find a second approximation.
Explain why the value of this second approximation will be greater than MMM, but less than 115(67111−1)\displaystyle \frac{1}{15}(671\sqrt{11} - 1)151(67111−1).
825 exam-style questions on OCR (MEI) A Level Maths 1.9 Calculus, covering 1.9.1 Gradient of a curve at a point, 1.9.2 Gradient as limit of chord gradient, 1.9.3 Derivative as gradient of tangent, 1.9.4 Sketch the gradient function, 1.9.5 Differentiate y = kx^n, 1.9.6 Second derivative as rate of change of gradient, 1.9.7 Stationary points: maxima and minima, 1.9.8 Increasing and decreasing functions, 1.9.9 Tangent and normal at a point, 1.9.10 Differentiate e^kx, a^kx and ln x (A-level only), 1.9.11 Differentiate trigonometric functions (A-level only), 1.9.12 Product rule (A-level only), 1.9.13 Quotient rule (A-level only), 1.9.14 Chain rule (A-level only), 1.9.15 Rates of change with the chain rule (A-level only), 1.9.16 Implicit differentiation (A-level only), 1.9.17 Concavity and the second derivative (A-level only), 1.9.18 Points of inflection (A-level only), 1.9.19 Integration as reverse of differentiation, 1.9.20 Integrate kx^n, 1.9.21 Constant of integration, 1.9.22 Indefinite and definite integrals, 1.9.23 Area between a graph and the x-axis, 1.9.24 Integrate e^kx, 1/x, sin kx, cos kx (A-level only), 1.9.25 Integration as the limit of a sum (A-level only), 1.9.26 Area between two curves (A-level only), 1.9.27 Integration by substitution (reverse chain rule) (A-level only), 1.9.28 Integration by substitution (other cases) (A-level only), 1.9.29 Integration by parts (A-level only), 1.9.30 Integration using partial fractions (A-level only), 1.9.31 Formulate first order differential equations (A-level only), 1.9.32 Solve first order differential equations (A-level only), and 1.9.33 Interpret solutions of differential equations (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.