Definite integrals questions with answers

Questions

1. Question 1Polynomials

   Evaluate $\int_{1}^{3} \left(x^{2} - 4\right) \, dx$.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{1}^{3} \left(x^{2} - 4\right) \, dx = \Big[\dfrac{1}{3}x^{3} - 4x\Big]_{1}^{3}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(\dfrac{1}{3}(3)^{3} - 4(3)\right) - \left(\dfrac{1}{3}(1)^{3} - 4(1)\right)$$= \Big(-3\Big) - \Big(-\dfrac{11}{3}\Big)$$= \dfrac{2}{3}$
2. Question 2Polynomials

   Evaluate $\int_{1}^{3} \left(-3x^{2} + 6x\right) \, dx$.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{1}^{3} \left(-3x^{2} + 6x\right) \, dx = \Big[-x^{3} + 3x^{2}\Big]_{1}^{3}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(-(3)^{3} + 3(3)^{2}\right) - \left(-(1)^{3} + 3(1)^{2}\right)$$= \Big(0\Big) - \Big(2\Big)$$= -2$
3. Question 3Powers of x and of (ax + b)

   Evaluate $\int_{1}^{2} (3x - 1)^{3} \, dx$.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{1}^{2} (3x - 1)^{3} \, dx = \Big[\dfrac{(3x - 1)^{4}}{12}\Big]_{1}^{2}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(\dfrac{(3(2) - 1)^{4}}{12}\right) - \left(\dfrac{(3(1) - 1)^{4}}{12}\right)$$= \Big(\dfrac{625}{12}\Big) - \Big(\dfrac{4}{3}\Big)$$= \dfrac{203}{4}$
4. Question 4Powers of x and of (ax + b)

   Evaluate $\int_{1}^{2} (3x + 1)^{2} \, dx$.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{1}^{2} (3x + 1)^{2} \, dx = \Big[\dfrac{(3x + 1)^{3}}{9}\Big]_{1}^{2}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(\dfrac{(3(2) + 1)^{3}}{9}\right) - \left(\dfrac{(3(1) + 1)^{3}}{9}\right)$$= \Big(\dfrac{343}{9}\Big) - \Big(\dfrac{64}{9}\Big)$$= 31$
5. Question 5Exponential, trigonometric and logarithmic

   Evaluate $\int_{0}^{\dfrac{\pi}{2}} 3\cos x \, dx$, giving your answer in exact form.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{0}^{\dfrac{\pi}{2}} 3\cos x \, dx = \Big[3\sin x\Big]_{0}^{\dfrac{\pi}{2}}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(3\sin (\dfrac{\pi}{2})\right) - \left(3\sin (0)\right)$$= \Big(3\Big) - \Big(0\Big)$$= 3$
6. Question 6Exponential, trigonometric and logarithmic

   Evaluate $\int_{0}^{2} 6e^{3x} \, dx$, giving your answer in exact form.

   Show the worked solution

   1. Step 1: Integrate (no + c is needed for a definite integral) and write the result in square brackets with the limits$\int_{0}^{2} 6e^{3x} \, dx = \Big[2e^{3x}\Big]_{0}^{2}$
   2. Step 2: Put in the upper limit, then subtract the value at the lower limit$= \left(2e^{3(2)}\right) - \left(2e^{3(0)}\right)$$= \Big(2e^{6}\Big) - \Big(2\Big)$$= 2(e^{6} - 1)$