Match the List-I with List-II
| List-I | List-II | ||
| Definite Integrals | Values | ||
| (A) | \( \int_{-1}^{1}\displaystyle |x|\,dx \) | (I) | 2 |
| (B) | \( \int_{-1}^{1}\displaystyle \sin^{3}x\cos^{4}x\ dx \) | (II) | 1 |
| (C) | \( \int_{-1}^{1}\displaystyle 3|x|^{2}\,dx \) | (III) | 0 |
| (D) | \( \int_{0}^{4}\displaystyle |x-1|\,dx \) | (IV) | 5 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
(A) \( \int_{-1}^{1}|x|\,dx \)
This matches (II) 1.
|x| is even, so
\( \displaystyle \int_{-1}^{1}|x|\,dx=2\int_{0}^{1}x\,dx \)
\( \displaystyle =2\left[\frac{x^{2}}{2}\right]_{0}^{1} \)
\( \displaystyle =1 \)
(B) \( \int_{-1}^{1}\sin^{3}x\cos^{4}x\ dx \)
This matches (III) 0.
Let \( f(x)=\sin^{3}x\cos^{4}x \). Then \( f(-x)=(-\sin x)^{3}(\cos x)^{4}=-f(x) \), so f is odd and its integral over [−1, 1] is 0.
(C) \( \int_{-1}^{1}3|x|^{2}\,dx \)
This matches (I) 2.
\( |x|^{2}=x^{2} \), which is even:
\( \displaystyle \int_{-1}^{1}3x^{2}\,dx=2\int_{0}^{1}3x^{2}\,dx \)
\( \displaystyle =2\left[x^{3}\right]_{0}^{1} \)
\( \displaystyle =2 \)
(D) \( \int_{0}^{4}|x-1|\,dx \)
This matches (IV) 5.
Splitting at x = 1, where x − 1 changes sign:
\( \displaystyle \int_{0}^{4}|x-1|\,dx=\int_{0}^{1}(1-x)\,dx+\int_{1}^{4}(x-1)\,dx \)
\( \displaystyle =\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4} \)
\( \displaystyle =\frac{1}{2}+\left(4-\left(-\frac{1}{2}\right)\right) \)
\( \displaystyle =\frac{1}{2}+\frac{9}{2} \)
\( \displaystyle =5 \)
So the matching is (A) - (II), (B) - (III), (C) - (I), (D) - (IV).
Shortcut: For (A) and (D), the graphs of |x| and |x − 1| are straight lines, so the integrals are areas of triangles:
\( \displaystyle \text{(A)}:\ 2\times\frac{1}{2}\times 1\times 1=1 \)
\( \displaystyle \text{(D)}:\ \frac{1}{2}\times 1\times 1+\frac{1}{2}\times 3\times 3=5 \)