Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Exercise 7.8 · Q1: ∫−11(x+1) dx\int_{-1}^{1}(x+1)\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫−11(x+1) dx.\int_{-1}^{1}(x+1)\,dx.
Exam answer

Let I=∫−11(x+1) dxI = \displaystyle\int_{-1}^{1}(x + 1)\,dx⁠.

I=[x22+x]−11=(12+1)−(12−1)=32+12\begin{aligned} I &= \left[\frac{x^2}{2} + x\right]_{-1}^{1} \\ &= \left(\frac12 + 1\right) - \left(\frac12 - 1\right) \\ &= \frac32 + \frac12 \end{aligned}

∴ I=2\therefore\ I = 2⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): if ff is continuous on [a,b][a, b] and F′=fF' = f⁠, then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)⁠. So we find one antiderivative of x+1x + 1⁠, then subtract its value at the lower limit −1-1 from its value at the upper limit 11⁠. Because x+1≥0x + 1 \ge 0 on [−1,1][-1, 1]⁠, the answer is also an area: a triangle we can measure directly.

Given, to find, and the plan
Given
The integrand f(x)=x+1f(x) = x + 1 on the interval [−1,1][-1, 1]⁠.
To find
The exact value of ∫−11(x+1) dx\int_{-1}^{1}(x+1)\,dx⁠.
Plan
(1) Check ff is continuous on [−1,1][-1, 1]⁠. (2) Find an antiderivative FF and differentiate it back. (3) Compute F(1)−F(−1)F(1) - F(-1)⁠. (4) Confirm with the triangle area and a numerical integration.

Basics you need

1Second fundamental theorem

∫abf(x) dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a) when ff is continuous on [a,b][a, b] and F′=fF' = f⁠. The constant C cancels in the subtraction, so it is left out. Lesson: definite integrals and the fundamental theorem

2Power rule

∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for n≠−1n \ne -1 (standard formula (i), Section 7.2); in particular ∫1 dx=x+C\int 1\,dx = x + C⁠. Basics: powers of x and the power rule

3Sum rule

∫(f+g) dx=∫f dx+∫g dx\int (f + g)\,dx = \int f\,dx + \int g\,dx (Property (III), Section 7.2.1): integrate term by term. Basics: integration undoes differentiation

4Area meaning

When f≥0f \ge 0 on [a,b][a, b]⁠, ∫abf(x) dx\int_a^b f(x)\,dx is the area between the graph and the x-axis from x=ax = a to x=bx = b (Section 7.8.1). Basics: area as a sum of thin strips

Solution

  1. Check continuity on [−1,1][-1, 1]

    f(x)=x+1f(x) = x + 1 is a polynomial. Polynomials are continuous at every real number, so ff is continuous on the whole closed interval [−1,1][-1, 1]⁠.

    Theorem 2 only applies when the integrand is defined and continuous on all of [a,b][a, b] (Remark (iv) after Theorem 2). For a polynomial there is nothing that can go wrong: no denominator, no square root, no logarithm.
    Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

  2. Find an antiderivative

    Integrate term by term (sum rule), using the power rule with n=1n = 1 for xx and ∫1 dx=x\int 1\,dx = x⁠:

    ∫(x+1) dx=x22+x+C.\int (x+1)\,dx = \frac{x^2}{2} + x + C.

    Take C=0C = 0⁠:

    F(x)=x22+x.F(x) = \frac{x^2}{2} + x.
    Any antiderivative works in Theorem 2, because a different choice F+CF + C adds the same C to F(b)F(b) and to F(a)F(a)⁠, and the two Cs cancel. So we pick the simplest one, with C=0C = 0⁠.
    Standard formula (i), Section 7.2, with n=1n = 1⁠: ∫x dx=x22+C\int x\,dx = \frac{x^2}{2} + C⁠. Basics: the power rule

  3. Check: differentiate FF back

    F′(x)=2x2+1=x+1=f(x).F'(x) = \frac{2x}{2} + 1 = x + 1 = f(x).
    An antiderivative is only correct if its derivative is exactly the integrand. Here it is, for every real xx⁠, so FF is a valid antiderivative on [−1,1][-1, 1]⁠.
    ddxx2=2x\frac{d}{dx}x^2 = 2x and ddxx=1\frac{d}{dx}x = 1⁠. Basics: integration undoes differentiation

  4. Evaluate F(1)−F(−1)F(1) - F(-1)

    Value at the upper limit:

    F(1)=122+1=12+1=32.F(1) = \frac{1^2}{2} + 1 = \frac12 + 1 = \frac32.

    Value at the lower limit (careful with the sign: (−1)2=1(-1)^2 = 1⁠):

    F(−1)=(−1)22+(−1)=12−1=−12.F(-1) = \frac{(-1)^2}{2} + (-1) = \frac12 - 1 = -\frac12.

    Subtract:

    ∫−11(x+1) dx=32−(−12)=32+12=2.\int_{-1}^{1}(x+1)\,dx = \frac32 - \left(-\frac12\right) = \frac32 + \frac12 = 2.
    The rule is "upper minus lower": F(b)−F(a)F(b) - F(a) with b=1b = 1 and a=−1a = -1⁠. Subtracting a negative number adds it, so the −12-\frac12 becomes +12+\frac12⁠.
    [F(x)]−11\big[F(x)\big]_{-1}^{1} means F(1)−F(−1)F(1) - F(-1)⁠. Lesson: the fundamental theorem

  5. See it as an area

    On [−1,1][-1, 1] we have x+1≥0x + 1 \ge 0⁠, so the integral is the area under the line y=x+1y = x + 1⁠. That region is a right triangle with vertices (−1,0)(-1, 0)⁠, (1,0)(1, 0) and (1,2)(1, 2)⁠: base 2, height 2.

    area=12×base×height=12×2×2=2.\text{area} = \frac12\times\text{base}\times\text{height} = \frac12\times 2\times 2 = 2.
    −1 1 2 x y y = x + 1 (1, 2) height 2 area 2
    The line y=x+1y = x + 1 meets the x-axis at x=−1x = -1⁠. Under it, from −1-1 to 11⁠, is a right triangle of base 2 and height 2, so its area is 2, the value of the integral.
    Two completely different methods (the fundamental theorem and school geometry) give the same number. That is exactly what Section 7.8 promises: for a non-negative integrand, the definite integral is the area under the graph.
    Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips
Result ∫−11(x+1) dx=2.\int_{-1}^{1}(x+1)\,dx = 2.

(Textbook answer: 2, the same.)

Check it with numbers

A computer's numerical integration of x+1x + 1 from −1-1 to 11 (adding very many thin strips, without using FF⁠) gives 2.00000002.0000000⁠. Another quick check: for a straight line, the integral equals the length of the interval times the value at its midpoint, 2×f(0)=2×1=22\times f(0) = 2\times 1 = 2⁠.

Common mistakes
  1. Getting F(−1)F(-1) wrong. (−1)22=+12\frac{(-1)^2}{2} = +\frac12⁠, not −12-\frac12⁠; then add −1-1 to get −12-\frac12⁠.
  2. Subtracting the wrong way. It is F(upper)−F(lower)=F(1)−F(−1)F(\text{upper}) - F(\text{lower}) = F(1) - F(-1)⁠.
  3. Losing the minus inside the bracket. 32−(−12)=2\frac32 - \left(-\frac12\right) = 2⁠, not 11⁠.
  4. Forgetting the ∫1 dx=x\int 1\,dx = x term. Integrating only the xx gives [x22]−11=0\big[\frac{x^2}{2}\big]_{-1}^{1} = 0⁠, which is wrong.
What to take away
  1. A definite integral is "antiderivative, then upper minus lower".
  2. No "+ C" is needed in a definite integral: it cancels.
  3. When the integrand is a straight line above the axis, the answer is a triangle or trapezium area: a free check.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 4Theorem 2, Section 7.8.3 (Lesson)
Polynomials are continuousStep 1Chapter 5, Section 5.2 (Example 14)
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Step 2Standard formula (i), Section 7.2 (Basics)
∫(f+g)=∫f+∫g\int (f + g) = \int f + \int gStep 2Property (III), Section 7.2.1
Differentiation undoes integrationStep 3Property (I), Section 7.2.1 (Basics)
Definite integral as area; triangle area 12bh\frac12 bhStep 5Section 7.8.1 (Basics)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →